Motion in One Dimension PYQs: MHT CET, JEE Main, NEET

Q1MHT CET 2026 · 15 MAY · SHIFT 2Easy

Distance, Displacement & Average Speed · Frame of reference, distance & displacement

A student runs along a circular track of diameter $80$ m. The distance travelled and the magnitude of displacement of the student when he covers $\left(\frac{3}{4}\right)^{\text{th}}$ of the circle is (in m)

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Answer: (a)

Radius $R = 80/2 = 40$ m.
Distance $= \frac{3}{4}(2\pi R) = \frac{3}{4}(80\pi) = 60\pi$ m.
After $\frac{3}{4}$ turn the angle swept is $270^\circ$ , so start and end radii are perpendicular: displacement $= \sqrt{R^2 + R^2} = R\sqrt{2} = 40\sqrt{2}$ m.
So the answer is (a) $60\pi, 40\sqrt{2}$ .

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Q2MHT CET 2026 · 15 MAY · SHIFT 2Medium

Distance, Displacement & Average Speed · Average speed: equal distances

A person travelling in a straight line moves with uniform velocity $V_1$ for a distance $x$ and with a uniform velocity $V_2$ for the next $\frac{3x}{2}$ distance. The average velocity in this motion is $\frac{50}{7}$ m/s. If $V_1$ is $5$ m/s then $V_2$ (in m/s) is

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Answer: (c)

Total distance $= x + \frac{3x}{2} = \frac{5x}{2}$ ; total time $= \frac{x}{5} + \frac{3x}{2V_2}$ .
$\frac{5x/2}{x/5 + 3x/(2V_2)} = \frac{50}{7} \Rightarrow \frac{1}{5} + \frac{3}{2V_2} = 0.35$
$\frac{3}{2V_2} = 0.15 \Rightarrow V_2 = 10$ m/s
So the answer is (c) $10$ .

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Q3MHT CET 2026 · 19 MAY · SHIFT 2Medium

Distance, Displacement & Average Speed · Average velocity vs average speed

The minute hand of a wall-clock is $10$ cm long. The average velocity of the tip of the minute hand between $6.00$ am to $7.30$ am is

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Answer: (b)

From 6:00 to 7:30 the minute hand makes $1\frac{1}{2}$ turns: tip moves from the 12 mark to the 6 mark.
Displacement $= 2r = 20$ cm; time $= 1.5 \times 3600 = 5400$ s.
Average velocity $= \frac{20}{5400} = \frac{1}{270}$ cm/s
So the answer is (b) $\frac{1}{270}$ cm/s.

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Q4NEET 2026 · RE-NEETMedium

Distance, Displacement & Average Speed · Average velocity vs average speed

Consider a particle moving along a straight line, whose position as a function of time is given by $s(t) = \alpha t^2 - \beta t + \gamma$ , where $\alpha = 1\text{ ms}^{-2}, \beta = 6\text{ ms}^{-1}$ and $\gamma = 5\text{ m}$ . The average speed of the particle, in $\text{ms}^{-1}$ from $t = 0$ to $t = 6\text{ s}$ is:

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Answer: (d)

$s = t^2 - 6t + 5$ , $v = 2t - 6 = 0$ at $t = 3$ s, so the particle turns back.
$s(0) = 5$ , $s(3) = -4$ , $s(6) = 5$ m.
Distance $= |{-4} - 5| + |5 - (-4)| = 9 + 9 = 18$ m
Average speed $= \frac{18}{6} = 3$ m/s, so the answer is (d) 3.

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Q5JEE Main 2025 · 2 APR · SHIFT 1Easy

Distance, Displacement & Average Speed · Average speed: equal distances

A person travelling on a straight line moves with a uniform velocity $v_1$ for a distance $x$ and with a uniform velocity $v_2$ for the next $\frac{3}{2}x$ distance. The average velocity in this motion is $\frac{50}{7}\text{ m/s}$ . If $v_1$ is $5\text{ m/s}$ then $v_2 =$ __________ $\text{m/s}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 10

Total distance $= x + \frac{3}{2}x = \frac{5x}{2}$ ; total time $= \frac{x}{5} + \frac{3x}{2v_2}$ .
$\frac{5x/2}{x/5 + 3x/(2v_2)} = \frac{50}{7} \Rightarrow \frac{1}{5} + \frac{3}{2v_2} = \frac{5}{2} \times \frac{7}{50} = 0.35$
$\frac{3}{2v_2} = 0.15 \Rightarrow v_2 = 10$
So $v_2 = 10$ m/s.

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Q6MHT CET 2024 · 10 MAY · SHIFT 2Easy

Distance, Displacement & Average Speed · Average speed: equal distances

A vehicle runs on a straight road of length 'L'. It travels half the distance with speed V and the remaining distance with speed $\frac{V}{3}$ . Its average speed is

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Answer: (c)

For equal distances, $v_{avg} = \frac{2v_1 v_2}{v_1 + v_2}$ .
$v_{avg} = \frac{2V(V/3)}{V + V/3} = \frac{2V^2/3}{4V/3}$
So the answer is (c) $\frac{V}{2}$ .

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Q7JEE Main 2024 · 9 APR · SHIFT 1Medium

Distance, Displacement & Average Speed · Average speed: mixed distance & time stages

A particle moving in a straight line covers half the distance with speed $6\text{ m/s}$ . The other half is covered in two equal time intervals with speeds $9\text{ m/s}$ and $15\text{ m/s}$ respectively. The average speed of the particle during the motion is:

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Answer: (d)

Let each half be $d$ . First half time $= \frac{d}{6}$ .
Second half: equal times $t$ at 9 and 15 m/s, so $9t + 15t = d \Rightarrow t = \frac{d}{24}$ , time $= 2t = \frac{d}{12}$ .
$v_{avg} = \frac{2d}{d/6 + d/12} = \frac{2d}{d/4} = 8$ m/s
So the answer is (d) 8 m/s.

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Q8JEE Main 2024 · 6 APR · SHIFT 1Medium

Distance, Displacement & Average Speed · Average speed: mixed distance & time stages

A train starting from rest first accelerates uniformly up to a speed of $80\text{ km/h}$ for time $t$ , then it moves with a constant speed for time $3t$ . The average speed of the train for this duration of journey will be (in $\text{km/h}$ ):

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Answer: (a)

Accelerating from rest to 80 km/h in time $t$ : distance $= \frac{0 + 80}{2}t = 40t$ .
Constant speed for $3t$ : distance $= 80 \times 3t = 240t$ .
$v_{avg} = \frac{40t + 240t}{4t} = 70$ km/h
So the answer is (a) 70.

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Q9NEET 2023Easy

Distance, Displacement & Average Speed · Average speed: equal distances

A vehicle travels half the distance with speed $v$ and the remaining distance with speed $2v$ . Its average speed is:

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Answer: (b)

For equal distances, $v_{avg} = \frac{2v_1 v_2}{v_1 + v_2}$ .
$v_{avg} = \frac{2(v)(2v)}{v + 2v} = \frac{4v^2}{3v}$
So the answer is (b) $\frac{4v}{3}$ .

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Q10JEE Main 2023 · 10 APR · SHIFT 2Easy

Distance, Displacement & Average Speed · Average speed: equal distances

A person travels $x$ distance with velocity $v_1$ and then $x$ distance with velocity $v_2$ in the same direction. The average velocity of the person is $v$ , then the relation between $v, v_1$ and $v_2$ will be.

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Answer: (d)

Same direction, so displacement $= 2x$ and time $= \frac{x}{v_1} + \frac{x}{v_2}$ .
$v = \frac{2x}{x/v_1 + x/v_2} \Rightarrow \frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2}$
So the answer is (d).

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Q11JEE Main 2023 · 30 JAN · SHIFT 2Easy

Distance, Displacement & Average Speed · Average speed: equal distances

A vehicle travels $4\text{ km}$ with speed of $3\text{ km/h}$ and another $4\text{ km}$ with speed of $5\text{ km/h}$ , then its average speed is

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Answer: (a)

Equal distances: $v_{avg} = \frac{2v_1 v_2}{v_1 + v_2}$ .
$v_{avg} = \frac{2(3)(5)}{3 + 5} = \frac{30}{8} = 3.75$ km/h
So the answer is (a) 3.75 km/h.

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Q12JEE Main 2023 · 25 JAN · SHIFT 1Easy

Distance, Displacement & Average Speed · Average speed: equal distances

A car travels a distance of ' $x$ ' with speed $v_1$ and then same distance ' $x$ ' with speed $v_2$ in the same direction. The average speed of the car is:

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Answer: (b)

Total distance $2x$ , total time $\frac{x}{v_1} + \frac{x}{v_2} = \frac{x(v_1 + v_2)}{v_1 v_2}$ .
$v_{avg} = \frac{2x}{t} = \frac{2v_1 v_2}{v_1 + v_2}$
So the answer is (b).

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Q13JEE Main 2023 · 1 FEB · SHIFT 1Medium

Distance, Displacement & Average Speed · Average speed: equal distances

An object moves with speed $v_1, v_2$ and $v_3$ along a line segment AB, BC and CD respectively as shown in figure. Where AB = BC and AD = 3AB, then average speed of the object will be:

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Answer: (d)

$AB = BC$ and $AD = 3AB$ , so $AB = BC = CD = d$ .
Total time $= \frac{d}{v_1} + \frac{d}{v_2} + \frac{d}{v_3} = \frac{d(v_2 v_3 + v_3 v_1 + v_1 v_2)}{v_1 v_2 v_3}$
$v_{avg} = \frac{3d}{t} = \frac{3v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$
So the answer is (d).

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Q14JEE Main 2023 · 30 JAN · SHIFT 1Medium

Distance, Displacement & Average Speed · Average speed: mixed distance & time stages

A horse rider covers half the distance with $5\text{ m/s}$ speed. The remaining part of the distance was travelled with speed $10\text{ m/s}$ for half the time and with speed $15\text{ m/s}$ for other half of the time. The mean speed of the rider averaged over the whole time of motion is $\frac{x}{7}\text{ m/s}$ . The value of $x$ is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 50

Second half in equal times at 10 and 15 m/s: its average speed $= \frac{10 + 15}{2} = 12.5$ m/s.
Now two equal distances at 5 and 12.5 m/s: $v_{avg} = \frac{2(5)(12.5)}{5 + 12.5} = \frac{125}{17.5} = \frac{50}{7}$ m/s
So $x = 50$ .

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Q15JEE Main 2022 · 28 JUN · SHIFT 2Easy

Distance, Displacement & Average Speed · Average speed: equal distances

A car covers AB distance with first one-third at velocity v1 ms ${}^{-1}$ , second one-third at v2 ms ${}^{-1}$ and last one-third at v3 ms ${}^{-1}$ . If v3 = 3v1, v2 = 2v1 and v1 = 11 ms ${}^{-1}$ then the average velocity of the car is _____________ ms______1.

Numerical answer type. Work it out, then open the answer.

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Answer: 18

Three equal parts $d$ : $v_{avg} = \frac{3d}{d/v_1 + d/v_2 + d/v_3}$ .
$v_1 = 11,\ v_2 = 22,\ v_3 = 33$ m/s
$\frac{1}{11} + \frac{1}{22} + \frac{1}{33} = \frac{6 + 3 + 2}{66} = \frac{1}{6}$
$v_{avg} = 3 \times 6 = 18$ m/s

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Q16NEET 2019Easy

Distance, Displacement & Average Speed · Average speed: equal distances

A person travelling in a straight line moves with a constant velocity $v_1$ for certain distance ' $x$ ' and with a constant velocity $v_2$ for next equal distance. The average velocity $v$ is given by the relation

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Answer: (c)

Total displacement $= 2x$ , total time $= \frac{x}{v_1} + \frac{x}{v_2}$ .
$v = \frac{2x}{x/v_1 + x/v_2} \Rightarrow \frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2}$
So the answer is (c).

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Q17NEET 2018Medium

Distance, Displacement & Average Speed · Average velocity vs average speed

A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field $\vec{E}$ . Due to the force q $\vec{E}$ , its velocity increases from 0 to 6 m s–1 in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively

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Answer: (b)

$a = \frac{6}{1} = 6$ m/s $^2$ . For 0–1 s: distance $= \frac{1}{2}(6)(1)^2 = 3$ m forward.
Field reversed, $a = -6$ : 1–2 s velocity $6 \to 0$ , 3 m forward; 2–3 s velocity $0 \to -6$ , 3 m backward.
Displacement $= 3 + 3 - 3 = 3$ m, distance $= 9$ m.
Average velocity $= \frac{3}{3} = 1$ m/s, average speed $= \frac{9}{3} = 3$ m/s
So the answer is (b).

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Q18NEET 2011 · AIPMT · MAINSEasy

Distance, Displacement & Average Speed · Average speed: equal distances

A particle covers half of its total distance with speed v1 and the rest half distance with speed v2. Its average speed during the complete journey is

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Answer: (c)

Let each half be $d$ . Total time $t = \frac{d}{v_1} + \frac{d}{v_2} = \frac{d(v_1 + v_2)}{v_1 v_2}$ .
$v_{avg} = \frac{2d}{t} = \frac{2v_1 v_2}{v_1 + v_2}$
So the answer is (c).

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Q19NEET 2007 · AIPMTEasy

Distance, Displacement & Average Speed · Average speed: equal distances

A car moves from X to Y with a uniform speed vu and returns to X with a uniform speed vd. The average speed for this round trip is

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Answer: (d)

XY and YX are equal distances $d$ , so total distance $= 2d$ .
Total time $= \frac{d}{v_u} + \frac{d}{v_d}$
$v_{avg} = \frac{2d}{d/v_u + d/v_d} = \frac{2v_u v_d}{v_u + v_d}$
So the answer is (d).

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Q20NEET 2006 · AIPMTEasy

Distance, Displacement & Average Speed · Average velocity vs average speed

A car runs at a constant speed on a circular track of radius 100 m, taking 62.8 seconds for every circular lap. The average velocity and average speed for each circular lap respectively is

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Answer: (c)

After one full lap the car returns to the start: displacement $= 0$ , so average velocity $= 0$ .
Average speed $= \frac{2\pi R}{T} = \frac{2 \times 3.14 \times 100}{62.8} = 10$ m/s
So the answer is (c) 0, 10 m/s.

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Q21NEET 1991 · AIPMTEasy

Distance, Displacement & Average Speed · Average speed: equal distances

A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of distance at speed $v$ . The average speed is 48 km/h. The value of $v$ is

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Answer: (b)

Equal distances: $v_{avg} = \frac{2v_1 v}{v_1 + v}$ .
$48 = \frac{2(40)v}{40 + v} \Rightarrow 1920 + 48v = 80v$
$32v = 1920 \Rightarrow v = 60$ km/h
So the answer is (b) 60 km/h.

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Q22NEET 1991 · AIPMTEasy

Distance, Displacement & Average Speed · Average speed: equal distances

A bus travelling the first one-third distance at a speed of 10 km/h, the next one-third at 20 km/h and at last one-third at 60 km/h. The average speed of the bus is

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Answer: (c)

Three equal distances $d$ : $v_{avg} = \frac{3d}{d/v_1 + d/v_2 + d/v_3}$ .
$\frac{1}{10} + \frac{1}{20} + \frac{1}{60} = \frac{6 + 3 + 1}{60} = \frac{1}{6}$
$v_{avg} = 3 \times 6 = 18$ km/h
So the answer is (c) 18 km/h.

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Q23NEET 1990 · AIPMTEasy

Distance, Displacement & Average Speed · Average speed: equal distances

A car covers the first half of the distance between two places at 40 km/h and another half at 60 km/h. The average speed of the car is

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Answer: (b)

Equal distances: $v_{avg} = \frac{2v_1 v_2}{v_1 + v_2}$ .
$v_{avg} = \frac{2(40)(60)}{100} = 48$ km/h
So the answer is (b) 48 km/h.

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Q24JEE Main 2026 · 28 JAN · SHIFT 2Medium

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

A particle starts moving from time $t = 0$ and its coordinate is given as $x(t) = 4t^3 - 3t$ A. The particle returns to its original position (origin) 0.866 units later

B. The particle is 1 unit away from origin at its turning point

C. Acceleration of the particle is non-negative

D. The particle is 0.5 units away from origin at its turning point

E. Particle never turns back as acceleration is non-negative

Choose the correct answer from the options given below:

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Answer: (b)

A: $x = 0 \Rightarrow 4t^2 = 3 \Rightarrow t = \dfrac{\sqrt{3}}{2} = 0.866$ , true.
Turning point: $v = 12t^2 - 3 = 0 \Rightarrow t = 0.5$ ; $x = 0.5 - 1.5 = -1$ , so B true, D false.
C: $a = 24t \ge 0$ for $t \ge 0$ , true.
E: the particle does turn back at $t = 0.5$ s, false.
So the answer is (b) A, B, C only.

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Q25MHT CET 2025 · 22 APR · SHIFT 2Medium

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

At any time ' $t$ ', the co-ordinates of moving particle are $x = at^2$ and $y = bt^2$ . The speed of the particle is

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Answer: (a)

$v_x = \dfrac{dx}{dt} = 2at$ , $v_y = \dfrac{dy}{dt} = 2bt$
Speed $= \sqrt{(2at)^2 + (2bt)^2} = 2t\sqrt{a^2 + b^2}$
So the answer is (a).

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Q26JEE Main 2025 · 3 APR · SHIFT 2Medium

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

A particle moves along the $x$ -axis and has its displacement $x$ varying with time t according to the equation:

$x = c_0\left(t^2 - 2\right) + c(t - 2)^2$

where $c_0$ and c are constants of appropriate dimensions.

Then, which of the following statements is correct?

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Answer: (a)

$v = \dfrac{dx}{dt} = 2c_0 t + 2c(t - 2)$
$a = \dfrac{dv}{dt} = 2c_0 + 2c = 2(c + c_0)$
Initial velocity $v(0) = -4c$ , so (d) is wrong.
So the answer is (a) $2(c + c_0)$ .

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Q27NEET 2025Medium

Velocity, Acceleration & Variable Motion · t(x) or v(x) given: a = v dv/dx

In some appropriate units, time $(t)$ and position $(x)$ relation of a moving particle is given by $t = x^2 + x$ . The acceleration of the particle is

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Answer: (d)

$\dfrac{dt}{dx} = 2x + 1 \Rightarrow v = \dfrac{1}{2x + 1}$
$\dfrac{dv}{dx} = -\dfrac{2}{(2x + 1)^2}$
$a = v\dfrac{dv}{dx} = -\dfrac{2}{(2x + 1)^3}$
So the answer is (d).

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Q28JEE Main 2025 · 29 JAN · SHIFT 2Difficult

Velocity, Acceleration & Variable Motion · Variable acceleration: integration

Two cars P and Q are moving on a road in the same direction. Acceleration of car P increases linearly with time whereas car Q moves with a constant acceleration. Both cars cross each other at time t = 0, for the first time. The maximum possible number of crossing(s) (including the crossing at t = 0) is ________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

Relative acceleration of P w.r.t. Q is linear in $t$ : $a_{rel} = k t + c$
Relative displacement from the crossing at $t = 0$ : $s_{rel} = ut + \dfrac{ct^2}{2} + \dfrac{kt^3}{6}$ , a cubic in $t$
A cubic has at most 3 roots, one of which is $t = 0$
So the maximum number of crossings is 3.

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Q29MHT CET 2024 · 11 MAY · SHIFT 2Easy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

Two cars start from a point at the same time in a straight line and their positions are represented by $x_1(t) = at + bt^2$ and $x_2(t) = Ft - t^2$ . At what time do the cars have the same velocity?

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Answer: (d)

$v_1 = \frac{dx_1}{dt} = a + 2bt$ , $v_2 = \frac{dx_2}{dt} = F - 2t$ .
$a + 2bt = F - 2t \Rightarrow 2t(1 + b) = F - a$
So the answer is (d) $t = \frac{F - a}{2(1 + b)}$ .

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Q30NEET 2024Easy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

A particle is moving along $x$ -axis with its position (x) varying with time $(t)$ as $x = \alpha t^4 + \beta t^2 + \gamma t + \delta$ . The ratio of its initial velocity to its initial acceleration, respectively, is:

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Answer: (d)

$v = 4\alpha t^3 + 2\beta t + \gamma$ , so at $t = 0$ : $v_0 = \gamma$ .
$a = 12\alpha t^2 + 2\beta$ , so at $t = 0$ : $a_0 = 2\beta$ .
So the answer is (d) $\gamma : 2\beta$ .

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Q31JEE Main 2024 · 29 JAN · SHIFT 2Easy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

A particle is moving in a straight line. The variation of position ' $x$ ' as a function of time ' $t$ ' is given as $x = \left(t^3 - 6t^2 + 20t + 15\right)\text{ m}$ . The velocity of the body when its acceleration becomes zero is:

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Answer: (c)

$v = 3t^2 - 12t + 20$ , $a = 6t - 12$ .
$a = 0 \Rightarrow t = 2$ s.
$v = 12 - 24 + 20 = 8$ m/s
So the answer is (c) 8 m/s.

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Q32JEE Main 2024 · 1 FEB · SHIFT 1Easy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

A particle is moving in one dimension (along $x$ axis) under the action of a variable force. Its initial position was $16\text{ m}$ right of origin. The variation of its position $(x)$ with time $(t)$ is given as $x = -3t^3 + 18t^2 + 16t$ , where $x$ is in $\text{m}$ and $t$ is in $\text{s}$ .

The velocity of the particle when its acceleration becomes zero is _________ $\text{m/s}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 52

$v = \dfrac{dx}{dt} = -9t^2 + 36t + 16$
$a = \dfrac{dv}{dt} = -18t + 36 = 0 \Rightarrow t = 2\text{ s}$
$v(2) = -36 + 72 + 16 = 52$
So the velocity is $52\text{ m/s}$ .

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Q33MHT CET 2024 · 16 MAY · SHIFT 1Medium

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

The co-ordinates of a moving particle at any time ' $t$ ' are given by $x = \alpha t^3$ and $y = \beta t^3$ where $\alpha$ and $\beta$ are constants. The speed of the particle at time ' $t$ ' is given by

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Answer: (d)

$v_x = 3\alpha t^2$ , $v_y = 3\beta t^2$
Speed $= \sqrt{v_x^2 + v_y^2} = 3t^2\sqrt{\alpha^2 + \beta^2}$
So the answer is (d).

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Q34JEE Main 2024 · 31 JAN · SHIFT 1Medium

Velocity, Acceleration & Variable Motion · t(x) or v(x) given: a = v dv/dx

The relation between time ' $t$ ' and distance ' $x$ ' is $t = \alpha x^2 + \beta x$ , where $\alpha$ and $\beta$ are constants. The relation between acceleration $(a)$ and velocity $(v)$ is:

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Answer: (c)

$\dfrac{dt}{dx} = 2\alpha x + \beta = \dfrac{1}{v}$
Differentiate w.r.t. $t$ : $2\alpha v = -\dfrac{1}{v^2}\dfrac{dv}{dt}$
$a = \dfrac{dv}{dt} = -2\alpha v^3$
So the answer is (c).

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Q35JEE Main 2024 · 6 APR · SHIFT 2Medium

Velocity, Acceleration & Variable Motion · t(x) or v(x) given: a = v dv/dx

A particle moves in a straight line so that its displacement $x$ at any time $t$ is given by $x^2 = 1 + t^2$ . Its acceleration at any time $t$ is $x^{-n}$ where $n =$ _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

Differentiate $x^2 = 1 + t^2$ : $xv = t \Rightarrow v = \dfrac{t}{x}$
Differentiate again: $v^2 + xa = 1 \Rightarrow a = \dfrac{1 - v^2}{x} = \dfrac{x^2 - t^2}{x^3}$
$x^2 - t^2 = 1$ , so $a = x^{-3}$
So $n = 3$ .

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Q36JEE Main 2024 · 1 FEB · SHIFT 2Medium

Velocity, Acceleration & Variable Motion · t(x) or v(x) given: a = v dv/dx

A particle initially at rest starts moving from reference point $x = 0$ along $x$ -axis, with velocity $v$ that varies as $v = 4\sqrt{x}\text{ m/s}$ . The acceleration of the particle is __________ $\text{ms}^{-2}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 8

$v^2 = 16x$
$a = v\dfrac{dv}{dx} = \dfrac{1}{2}\dfrac{d(v^2)}{dx} = \dfrac{16}{2}$
So $a = 8\text{ m/s}^2$ .

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Q37MHT CET 2023 · 13 MAY · SHIFT 1Easy

Velocity, Acceleration & Variable Motion · Speed, velocity & acceleration: concepts

Which one of the following statements is Wrong?

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Answer: (b)

(a) True: at the top of a vertical throw $v = 0$ but $a = g$ .
(b) Wrong: constant velocity means constant magnitude and direction, so speed cannot change.
(c) True: uniform circular motion has constant speed but changing velocity.
(d) True: in projectile motion $a = g$ is constant but the velocity direction changes.
So the answer is (b).

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Q38MHT CET 2023 · 14 MAY · SHIFT 2Easy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

The position ' $x$ ' of a particle varies with a time as $x = at^2 - bt^3$ where ' $a$ ' and ' $b$ ' are constants. The acceleration of the particle will be zero at

Show answer and solution

Answer: (c)

$v = \frac{dx}{dt} = 2at - 3bt^2$ , $a_{p} = \frac{dv}{dt} = 2a - 6bt$ .
$2a - 6bt = 0 \Rightarrow t = \frac{a}{3b}$
So the answer is (c) $\frac{a}{3b}$ .

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Q39MHT CET 2023 · 11 MAY · SHIFT 2Easy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

Two cars A and B start from a point at the same time in a straight line and their positions are represented by $R_A(t) =$ at $+bt^2$ and $R_B(t) = xt - t^2$ . At what time do the cars have same velocity?

Show answer and solution

Answer: (a)

$v_A = a + 2bt$ , $v_B = x - 2t$ .
$a + 2bt = x - 2t \Rightarrow 2t(b + 1) = x - a$
So the answer is (a) $t = \frac{x - a}{2(b + 1)}$ .

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Q40NEET 2023Easy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

The position of a particle is given by

$\vec{r}(t) = 4t\hat{i} + 2t^2\hat{j} + 5\hat{k}$

where $t$ is in seconds and $r$ in metre. Find the magnitude and direction of velocity $v(t)$ , at $t = 1\text{ s}$ , with respect to $x$ -axis

Show answer and solution

Answer: (a)

$\vec{v} = \frac{d\vec{r}}{dt} = 4\hat{i} + 4t\hat{j}$ ; at $t = 1$ s, $\vec{v} = 4\hat{i} + 4\hat{j}$ .
$|\vec{v}| = \sqrt{16 + 16} = 4\sqrt{2}$ m/s
$\tan\theta = \frac{4}{4} = 1 \Rightarrow \theta = 45^\circ$ with the $x$ -axis
So the answer is (a).

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Q41JEE Main 2023 · 15 APR · SHIFT 1Easy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

The position of a particle related to time is given by $x = \left(5t^2 - 4t + 5\right)\text{ m}$ . The magnitude of velocity of the particle at $t = 2\text{ s}$ will be:

Show answer and solution

Answer: (b)

$v = \frac{dx}{dt} = 10t - 4$
At $t = 2$ s: $v = 20 - 4 = 16$ m/s
So the answer is (b) $16\text{ ms}^{-1}$ .

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Q42JEE Main 2023 · 13 APR · SHIFT 2Easy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

The distance travelled by an object in time $t$ is given by $s = (2.5)t^2$ . The instantaneous speed of the object at $t = 5\text{ s}$ will be:

Show answer and solution

Answer: (d)

$v = \frac{ds}{dt} = 2 \times 2.5\, t = 5t$
At $t = 5$ s: $v = 25$ m/s
So the answer is (d) $25\text{ ms}^{-1}$ .

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Q43JEE Main 2023 · 25 JAN · SHIFT 2Easy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

The distance travelled by a particle is related to time t as $x = 4t^2$ . The velocity of the particle at t=5s is :-

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Answer: (d)

$v = \frac{dx}{dt} = 8t$
At $t = 5$ s: $v = 40$ m/s
So the answer is (d) $40\text{ ms}^{-1}$ .

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Q44JEE Main 2022 · 25 JUN · SHIFT 2Easy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

Two buses P and Q start from a point at the same time and move in a straight line and their positions are represented by $X_P(t) = \alpha t + \beta t^2$ and $X_Q(t) = ft - t^2$ . At what time, both the buses have same velocity?

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Answer: (d)

$v_P = \alpha + 2\beta t$ , $v_Q = f - 2t$ .
$\alpha + 2\beta t = f - 2t \Rightarrow 2t(1 + \beta) = f - \alpha$
So the answer is (d) $t = \frac{f - \alpha}{2(1 + \beta)}$ .

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Q45JEE Main 2022 · 25 JUL · SHIFT 2Easy

Velocity, Acceleration & Variable Motion · t(x) or v(x) given: a = v dv/dx

A particle is moving in a straight line such that its velocity is increasing at 5 ms ${}^{-1}$ per meter. The acceleration of the particle is _____________ ms ${}^{-2}$ at a point where its velocity is 20 ms ${}^{-1}$ .

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 100

Velocity increases at $5$ m/s per metre, so $\dfrac{dv}{dx} = 5\text{ s}^{-1}$
$a = v\dfrac{dv}{dx} = 20 \times 5$
So $a = 100\text{ m/s}^2$ .

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Q46JEE Main 2022 · 29 JUL · SHIFT 1Medium

Velocity, Acceleration & Variable Motion · t(x) or v(x) given: a = v dv/dx

If $t = \sqrt{x} + 4$ , then $\left(\frac{dx}{dt}\right)_{t=4}$ is:

Show answer and solution

Answer: (b)

$\sqrt{x} = t - 4 \Rightarrow x = (t - 4)^2$
$\dfrac{dx}{dt} = 2(t - 4)$
At $t = 4$ : $\dfrac{dx}{dt} = 0$
So the answer is (b) zero.

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Q47JEE Main 2021 · 25 JUL · SHIFT 2Easy

Velocity, Acceleration & Variable Motion · Variable acceleration: integration

The instantaneous velocity of a particle moving in a straight line is given as $V = \alpha t + \beta t^2$ , where $\alpha$ and $\beta$ are constants. The distance travelled by the particle between 1s and 2s is:

Show answer and solution

Answer: (b)

Distance $= \int_1^2 (\alpha t + \beta t^2)\,dt$
$= \dfrac{\alpha}{2}(2^2 - 1^2) + \dfrac{\beta}{3}(2^3 - 1^3)$
$= \dfrac{3}{2}\alpha + \dfrac{7}{3}\beta$
So the answer is (b).

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Q48JEE Main 2021 · 17 MAR · SHIFT 2Easy

Velocity, Acceleration & Variable Motion · Variable acceleration: integration

The velocity of a particle is v = v0 + gt + Ft2. Its position is x = 0 at t = 0; then its displacement after time (t = 1) is:

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Answer: (b)

$x = \int_0^1 (v_0 + gt + Ft^2)\,dt$ with $x(0) = 0$
$x = \left[v_0 t + \dfrac{g t^2}{2} + \dfrac{F t^3}{3}\right]_0^1$
$x = v_0 + \dfrac{g}{2} + \dfrac{F}{3}$
So the answer is (b).

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Q49JEE Main 2021 · 25 JUL · SHIFT 2Medium

Velocity, Acceleration & Variable Motion · t(x) or v(x) given: a = v dv/dx

The relation between time t and distance x for a moving body is given as t = mx2 + nx, where m and n are constants. The retardation of the motion is: (When v stands for velocity)

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Answer: (a)

$\dfrac{dt}{dx} = 2mx + n = \dfrac{1}{v}$
Differentiate w.r.t. $t$ : $2mv = -\dfrac{1}{v^2}\dfrac{dv}{dt} \Rightarrow a = -2mv^3$
Retardation $= 2mv^3$
So the answer is (a).

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Q50JEE Main 2021 · 27 AUG · SHIFT 1Medium

Velocity, Acceleration & Variable Motion · t(x) or v(x) given: a = v dv/dx

If the velocity of a body related to displacement x is given by $v = \sqrt{5000 + 24x}$ m/s, then the acceleration of the body is .................... m/s2.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 12

$v^2 = 5000 + 24x$
$a = \dfrac{1}{2}\dfrac{d(v^2)}{dx} = \dfrac{24}{2}$
So $a = 12\text{ m/s}^2$ .

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Q51JEE Main 2020 · 9 JAN · SHIFT 1Difficult

Velocity, Acceleration & Variable Motion · t(x) or v(x) given: a = v dv/dx

The distance x covered by a particle in one dimensional motion varies with time t as x2 = at2 + 2bt + c. If the acceleration of the particle depends on x as x–n, where n is an integer, the value of n is __________

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 3

Differentiate: $2xv = 2at + 2b \Rightarrow xv = at + b$
Differentiate again: $v^2 + xa_c = a \Rightarrow a_c = \dfrac{a - v^2}{x}$
$a_c = \dfrac{a x^2 - (at + b)^2}{x^3} = \dfrac{ac - b^2}{x^3}$
So $a_c \propto x^{-3}$ and $n = 3$ .

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Q52JEE Main 2019 · 9 APR · SHIFT 2Easy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

The position of a particle as a function of time t, is given by

x(t) = at + bt2 – ct3

where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be:

Show answer and solution

Answer: (c)

$v = a + 2bt - 3ct^2$ , acceleration $= 2b - 6ct$ .
Zero acceleration at $t = \frac{b}{3c}$ .
$v = a + 2b\frac{b}{3c} - 3c\frac{b^2}{9c^2} = a + \frac{2b^2}{3c} - \frac{b^2}{3c}$
So the answer is (c) $a + \frac{b^2}{3c}$ .

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Q53JEE Main 2019 · 9 APR · SHIFT 2Easy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

The position vector of a particle changes with time according to the relation

$\vec{r}(t) = 15t^2\hat{i} + (4 - 20t^2)\hat{j}$

What is the magnitude of the acceleration at t = 1?

Show answer and solution

Answer: (a)

$\vec{v} = \dfrac{d\vec{r}}{dt} = 30t\,\hat{i} - 40t\,\hat{j}$
$\vec{a} = \dfrac{d\vec{v}}{dt} = 30\hat{i} - 40\hat{j}$ (constant)
$|\vec{a}| = \sqrt{30^2 + 40^2} = 50$
So the answer is (a) 50.

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Q54JEE Main 2019 · 12 APR · SHIFT 2Medium

Velocity, Acceleration & Variable Motion · Variable acceleration: integration

A particle is moving with speed v = b $\sqrt{x}$ along positive x-axis. Calculate the speed of the particle at time t = $\tau$ (assume that the particle is at origin t = 0)

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Answer: (c)

$\dfrac{dx}{\sqrt{x}} = b\,dt \Rightarrow 2\sqrt{x} = bt$ (since $x = 0$ at $t = 0$ )
$x = \dfrac{b^2 t^2}{4}$
$v = \dfrac{dx}{dt} = \dfrac{b^2 t}{2}$ ; at $t = \tau$ , $v = \dfrac{b^2\tau}{2}$
So the answer is (c).

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Q55NEET 2016 · PHASE 2Easy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

Two cars P and Q start from a point at the same time in a straight line and their positions are represented by

xP(t) = (at + bt2) and xQ(t) = (ft $-$ t2).

At what time do the cars have the same velocity?

Show answer and solution

Answer: (d)

$v_P = a + 2bt$ , $v_Q = f - 2t$ .
$a + 2bt = f - 2t \Rightarrow 2t(1 + b) = f - a$
So the answer is (d) $t = \frac{f - a}{2(1 + b)}$ .

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Q56NEET 2016 · PHASE 1Easy

Velocity, Acceleration & Variable Motion · Variable acceleration: integration

If the velocity of a particle is v = At + Bt2, where A and B are constants, then the distance travelled by it between 1 s and 2 s is

Show answer and solution

Answer: (a)

$v > 0$ for $t$ between 1 s and 2 s (A, B positive), so distance $= \int_1^2 v\,dt$
$\int_1^2 (At + Bt^2)\,dt = \dfrac{A}{2}(4 - 1) + \dfrac{B}{3}(8 - 1)$
$= \dfrac{3}{2}A + \dfrac{7}{3}B$
So the answer is (a).

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Q57NEET 2015 · AIPMT · CANCELLED PAPERMedium

Velocity, Acceleration & Variable Motion · t(x) or v(x) given: a = v dv/dx

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to $v(x) = \beta x^{-2n}$ , where $\beta$ and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by

Show answer and solution

Answer: (d)

$a = v\dfrac{dv}{dx}$
$\dfrac{dv}{dx} = -2n\beta x^{-2n-1}$
$a = \beta x^{-2n} \cdot (-2n\beta x^{-2n-1}) = -2n\beta^2 x^{-4n-1}$
So the answer is (d).

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Q58NEET 2013Medium

Velocity, Acceleration & Variable Motion · t(x) or v(x) given: a = v dv/dx

The displacement 'x' (in meter) of a particle of mass 'm' (in kg) moving in one dimension under the action of a force, is related to time 't' (in sec) by t = $\sqrt{x} + 3$ . The displacement of the particle when its velocity is zero, will be

Show answer and solution

Answer: (b)

$\sqrt{x} = t - 3 \Rightarrow x = (t - 3)^2$
$v = \dfrac{dx}{dt} = 2(t - 3) = 0 \Rightarrow t = 3\text{ s}$
$x(3) = 0$
So the answer is (b) 0 m.

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Q59NEET 2012 · AIPMT · PRELIMSEasy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

The motion of a particle along a straight line is described by equation x = 8 + 12t $-$ t3 where x is in metre and t in second. The retardation of the particle when its velocity becomes zero is

Show answer and solution

Answer: (d)

$v = \frac{dx}{dt} = 12 - 3t^2 = 0 \Rightarrow t = 2$ s.
$a = \frac{dv}{dt} = -6t = -12$ m/s $^2$ at $t = 2$ s.
So the retardation is (d) 12 m s $^{-2}$ .

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Q60JEE Main 2011 · AIEEEMedium

Velocity, Acceleration & Variable Motion · Variable acceleration: integration

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by:

$\frac{dv}{dt} = -2.5\sqrt{v}$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be:

Show answer and solution

Answer: (a)

$\dfrac{dv}{\sqrt{v}} = -2.5\,dt$
$\int_{6.25}^{0} v^{-1/2}dv = -2.5\,t \Rightarrow -2\sqrt{6.25} = -2.5\,t$
$t = \dfrac{2 \times 2.5}{2.5} = 2\text{ s}$
So the answer is (a) 2 s.

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Q61NEET 2010 · AIPMT · PRELIMSMedium

Velocity, Acceleration & Variable Motion · t(x) or v(x) given: a = v dv/dx

A particle moves a distance x in time t according to equation x = (t + 5) ${}^{-1}$ . The acceleration of particle is proportional to

Show answer and solution

Answer: (a)

$x = (t + 5)^{-1}$ , $v = -(t + 5)^{-2}$ , $a = 2(t + 5)^{-3}$
$|v|^{3/2} = (t + 5)^{-3}$
So $a \propto v^{3/2}$ ; the answer is (a).

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Q62JEE Main 2007 · AIEEEEasy

Velocity, Acceleration & Variable Motion · Variable acceleration: integration

The velocity of a particle is v = v0 + gt + ft2. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is

Show answer and solution

Answer: (c)

$x = \int_0^1 (v_0 + gt + ft^2)\,dt$ with $x(0) = 0$
$x = v_0 t + \dfrac{g t^2}{2} + \dfrac{f t^3}{3}$ at $t = 1$
$x = v_0 + \dfrac{g}{2} + \dfrac{f}{3}$
So the answer is (c).

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Q63NEET 2007 · AIPMTMedium

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

The position x of a particle with respect to time t along x-axis is given by x = 9t2 $-$ t3 where x is in metres and t in seconds. What will be the position of this particle when it achieves maximum speed along the + x direction?

Show answer and solution

Answer: (a)

$v = \dfrac{dx}{dt} = 18t - 3t^2$
Speed is maximum when $a = \dfrac{dv}{dt} = 18 - 6t = 0 \Rightarrow t = 3\text{ s}$
$x(3) = 9(9) - 27 = 54\text{ m}$
So the answer is (a) 54 m.

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Q64NEET 2007 · AIPMTMedium

Velocity, Acceleration & Variable Motion · Variable acceleration: integration

A particle moving along x-axis has acceleration f, at time t, given by f = f0 $\left(1 - \frac{t}{T}\right),$ where f0 and T are constants. The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0, the particle's velocity (vx) is

Show answer and solution

Answer: (c)

$f = 0$ when $t = T$
$v = \int_0^T f_0\left(1 - \dfrac{t}{T}\right)dt = f_0\left[t - \dfrac{t^2}{2T}\right]_0^T$
$v = f_0 T - \dfrac{f_0 T}{2} = \dfrac{1}{2}f_0 T$
So the answer is (c) $\dfrac{1}{2}f_0 T$ (positive).

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Q65NEET 2006 · AIPMTMedium

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle from O is given by x = 40 + 12t $-$ t3. How long would the particle travel before coming to rest?

Show answer and solution

Answer: (a)

$v = \dfrac{dx}{dt} = 12 - 3t^2 = 0 \Rightarrow t = 2\text{ s}$
$x(0) = 40\text{ m}$ , $x(2) = 40 + 24 - 8 = 56\text{ m}$
$v > 0$ throughout, so distance $= 56 - 40 = 16\text{ m}$
So the answer is (a) 16 m.

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Q66JEE Main 2006 · AIEEEMedium

Velocity, Acceleration & Variable Motion · Variable acceleration: integration

A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity 'v' that varies as $v = \alpha\sqrt{x}$ . The displacement of the particle varies with time as

Show answer and solution

Answer: (a)

$\dfrac{dx}{\sqrt{x}} = \alpha\,dt \Rightarrow 2\sqrt{x} = \alpha t$
$x = \dfrac{\alpha^2 t^2}{4}$
So $x \propto t^2$ ; the answer is (a).

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Q67NEET 2005 · AIPMTMedium

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

The displacement x of a particle varies with time t as x = ae $-\alpha$ t + be $\beta$ t, where a, b, $\alpha$ and $\beta$ are positive constants. The velocity of the particle will

Show answer and solution

Answer: (d)

$v = -a\alpha e^{-\alpha t} + b\beta e^{\beta t}$
$\dfrac{dv}{dt} = a\alpha^2 e^{-\alpha t} + b\beta^2 e^{\beta t} > 0$ for all $t$
Acceleration is always positive, so $v$ keeps increasing.
So the answer is (d).

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Q68JEE Main 2005 · AIEEEMedium

Velocity, Acceleration & Variable Motion · t(x) or v(x) given: a = v dv/dx

The relation between time t and distance x is t = ax2 + bx where a and b are constants. The acceleration is

Show answer and solution

Answer: (d)

$\dfrac{dt}{dx} = 2ax + b = \dfrac{1}{v}$
Differentiate w.r.t. $t$ : $2av = -\dfrac{1}{v^2}\dfrac{dv}{dt}$
$a_{cc} = \dfrac{dv}{dt} = -2av^3$
So the answer is (d).

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Q69JEE Main 2003 · AIEEEEasy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

The co-ordinates of a moving particle at any time 't' are given by x = $\alpha$ t3 and y = βt3. The speed of the particle at time 't' is given by

Show answer and solution

Answer: (b)

$v_x = \dfrac{dx}{dt} = 3\alpha t^2$ , $v_y = \dfrac{dy}{dt} = 3\beta t^2$
Speed $= \sqrt{v_x^2 + v_y^2} = 3t^2\sqrt{\alpha^2 + \beta^2}$
So the answer is (b) $3t^2\sqrt{\alpha^2 + \beta^2}$ .

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Q70NEET 2000 · AIPMTEasy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

Motion of a particle is given by equation s = (3t3 + 7t2 + 14t + 8) m. The value of acceleration of the particle at t = 1 sec is

Show answer and solution

Answer: (b)

$v = \frac{ds}{dt} = 9t^2 + 14t + 14$
$a = \frac{dv}{dt} = 18t + 14$
At $t = 1$ s: $a = 18 + 14 = 32$ m/s $^2$ , so the answer is (b).

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Q71NEET 1997 · AIPMTEasy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

The position $x$ of a particle varies with time ( $t$ ) as $x = at^2 - bt^3$ . The acceleration will be zero at time $t$ is equal to

Show answer and solution

Answer: (a)

$v = 2at - 3bt^2$ , acceleration $= 2a - 6bt$ .
$2a - 6bt = 0 \Rightarrow t = \frac{a}{3b}$
So the answer is (a) $\frac{a}{3b}$ .

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Q72NEET 1995 · AIPMTMedium

Velocity, Acceleration & Variable Motion · Variable acceleration: integration

The acceleration of a particle is increasing linearly with time $t$ as $bt$ . The particle starts from origin with an initial velocity $v_0$ . The distance travelled by the particle in time $t$ will be

Show answer and solution

Answer: (c)

$a = bt \Rightarrow v = v_0 + \dfrac{bt^2}{2}$
$x = \int_0^t v\,dt = v_0 t + \dfrac{bt^3}{6}$
So the answer is (c) $v_0 t + \dfrac{1}{6}bt^3$ .

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Q73NEET 1994 · AIPMTEasy

Velocity, Acceleration & Variable Motion · x(t) given: find v and a

A particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^3 - 6t^2 + 3t + 4)$ metres. The velocity when the acceleration is zero is

Show answer and solution

Answer: (c)

$v = 3t^2 - 12t + 3$ , $a = 6t - 12$ .
$a = 0 \Rightarrow t = 2$ s.
$v = 3(4) - 24 + 3 = -9$ m/s
So the answer is (c) $-9$ m/s.

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Q74MHT CET 2026 · 13 APR · SHIFT 2Easy

Graphs of Motion · Velocity-time graphs: distance & displacement

The velocity time graph of a body moving in a straight line is shown in a figure. The ratio of displacement to distance travelled by the body in time $0$ to $10$ s is

Show answer and solution

Answer: (d)

Areas: $0$ - $2$ s: $8 \times 2 = 16$ ; $2$ - $4$ s: $-4 \times 2 = -8$ ; $4$ - $8$ s: $4 \times 4 = 16$ ; $8$ - $10$ s: $-4 \times 2 = -8$ (m)
Displacement $= 16 - 8 + 16 - 8 = 16\text{ m}$
Distance $= 16 + 8 + 16 + 8 = 48\text{ m}$
Ratio $= 16 : 48 = 1 : 3$ , so the answer is (d).

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Q75JEE Main 2026 · 5 APR · SHIFT 2Easy

Graphs of Motion · Velocity-time graphs: distance & displacement

The velocity $(v)$ versus time $(t)$ plot of a particle is shown in the figure, for a time interval of 40 s. The total distance travelled by the particle and the average velocity during this period are, respectively

______.

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Answer: (c)

The graph has four triangles, each of base $10\text{ s}$ and height $5\text{ m/s}$ : area $= \frac{1}{2}\times 10\times 5 = 25\text{ m}$ each.
Distance = sum of magnitudes $= 4\times 25 = 100\text{ m}$
Displacement = $25 + 25 - 25 - 25 = 0$ , so average velocity $= 0$
So the answer is (c) 100 m and zero.

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Q76JEE Main 2026 · 24 JAN · SHIFT 2Medium

Graphs of Motion · Velocity-displacement graphs

The velocity $(v)$ - Distance $(x)$ graph is shown in figure. Which graph represents acceleration(a) versus distance ( $x$ ) variation of this system?

Show answer and solution

Answer: (d)

From the graph $v = v_0 - kx$ (straight line, slope $-k$ ).
$a = v\frac{dv}{dx} = (v_0 - kx)(-k) = -kv_0 + k^2x$
So $a$ - $x$ is a straight line with negative intercept $-kv_0$ and positive slope $k^2$ , crossing zero where $v = 0$ .
So the answer is (d).

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Q77MHT CET 2025 · 21 APR · SHIFT 1Easy

Graphs of Motion · Velocity-time graphs: distance & displacement

The velocity time graph of a body moving in a straight line is shown in figure. The ratio of displacement to distance travelled by the body in time 0 to 8 s is

Show answer and solution

Answer: (c)

Read the graph (not to scale) as: $+3$ m/s for $0$ - $2$ s, $-2$ m/s for $2$ - $3$ s, $+2$ m/s for $3$ - $5$ s, $-2$ m/s for $5$ - $6$ s, $+2$ m/s for $6$ - $8$ s.
Positive area $= 6 + 4 + 4 = 14\text{ m}$ ; negative area $= 2 + 2 = 4\text{ m}$
Displacement $= 14 - 4 = 10\text{ m}$ ; distance $= 14 + 4 = 18\text{ m}$
Ratio $= 10 : 18 = 5 : 9$ , so the answer is (c).

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Q78JEE Main 2025 · 28 JAN · SHIFT 2Easy

Graphs of Motion · Velocity-time graphs: distance & displacement

The velocity-time graph of an object moving along a straight line is shown in the figure. What is the distance covered by the object between $t = 0$ to $t = 4\text{ s}$ ?

Show answer and solution

Answer: (b)

Distance = area under the $v$ - $t$ graph from 0 to 4 s.
0 to 2 s (triangle): $\frac{1}{2}\times 2\times 10 = 10\text{ m}$
2 to 4 s (rectangle): $2\times 10 = 20\text{ m}$
Total $= 30\text{ m}$ . So the answer is (b).

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Q79JEE Main 2025 · 23 JAN · SHIFT 1Easy

Graphs of Motion · Velocity-time graphs: distance & displacement

The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is ̱_______ km.

Show answer and solution

Answer: (a)

Distance = area under the $v$ - $t$ graph.
0 to 2 s (trapezium): $\frac{1}{2}(200 + 400)\times 2 = 600\text{ m}$
2 s to 30.5 s (rectangle): $400\times 28.5 = 11400\text{ m}$
Total $= 12000\text{ m} = 12\text{ km}$ . So the answer is (a).

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Q80JEE Main 2025 · 3 APR · SHIFT 1Easy

Graphs of Motion · Identifying and matching graphs

Which of the following curves possibly represent one-dimensional motion of a particle?

Choose the correct answer from the options given below:

Show answer and solution

Answer: (c)

(A) Phase increasing linearly with time ( $\phi = \omega t + \phi_0$ ) is possible, e.g. SHM.
(B) A closed $v$ - $x$ loop (ellipse/circle) is SHM along a line; clockwise sense is correct ( $v > 0$ while $x$ increases). Possible.
(C) A $v$ - $t$ circle gives two velocities at one instant and needs time to go backward. Not possible.
(D) Total distance never decreasing with time is possible.
So A, B and D only. So the answer is (c).

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Q81JEE Main 2025 · 4 APR · SHIFT 2Medium

Graphs of Motion · Position-time graphs

The displacement x versus time graph is shown below.

(A) The average velocity during 0 to 3 s is $10\text{ m/s}$

(B) The average velocity during 3 to 5 s is $0\text{ m/s}$

(C) The instantaneous velocity at $t = 2\text{ s}$ is $5\text{ m/s}$

(D) The average velocity during 5 to 7 s and instantaneous velocity at $t = 6.5\text{ s}$ are equal (E) The average velocity from $t = 0$ to $t = 9\text{ s}$ is zero

Choose the correct answer from the options given below:

Show answer and solution

Answer: (d)

(A) $\bar v_{0\to 3} = \frac{5 - 0}{3} = \frac{5}{3}\text{ m/s}$ , not 10. False.
(B) $x(3) = x(5) = 5\text{ m}$ , so $\bar v = 0$ . True.
(C) At 2 s the body is on line AC: slope $= \frac{5 - (-5)}{3 - 1} = 5\text{ m/s}$ . True.
(D) $\bar v_{5\to 7} = \frac{0 - 5}{2} = -2.5\text{ m/s}$ ; at 6.5 s slope of EF $= \frac{0 - 10}{1} = -10\text{ m/s}$ . Not equal. False.
(E) $x(9) = x(0) = 0$ , so $\bar v = 0$ . True.
Correct: (B), (C), (E). So the answer is (d).

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Q82MHT CET 2025 · 23 APR · SHIFT 1Medium

Graphs of Motion · Velocity-time graphs: slope, a-t graphs

The acceleration (a) - time (T) graph for the body starting from rest is given below. The maximum speed of the body is

Show answer and solution

Answer: (a)

Area under $a$ - $t$ graph = change in velocity.
$a > 0$ up to 10 s, so speed is maximum at $t = 10\text{ s}$ .
$v_{max} = 0 + \frac{1}{2}\times 10\times 8 = 40\text{ m/s}$ . So the answer is (a).

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Q83MHT CET 2024 · 9 MAY · SHIFT 2Easy

Graphs of Motion · Position-time graphs

The following figures show the variation of displacement with time of a particular object.

Show answer and solution

Answer: (a)

Figure (a): $x$ constant, slope zero, so the object is at rest.
Figure (b): straight line with positive constant slope, so uniform velocity in $+x$ direction.
Figure (c): constant negative slope (uniform velocity in $-x$ ), not rest; (d) changes direction.
So the answer is (a).

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Q84MHT CET 2024 · 11 MAY · SHIFT 1Easy

Graphs of Motion · Velocity-time graphs: distance & displacement

For the velocity-time graph shown in the figure below, the distance covered by the body in last two second of its motion is ' $S_1$ '. What is the ratio of ' $S_1$ ' to the total distance covered by it

Show answer and solution

Answer: (b)

Distance = area under the $v$ - $t$ graph. Read the time axis as A at $1\text{ s}$ , B at $3\text{ s}$ , C at $5\text{ s}$ , D at $7\text{ s}$ (the last label '4' is a misprint for 7).
Total distance = trapezium area $= \frac{1}{2}(2 + 6)\times 10 = 40\text{ m}$
Last two seconds (5 s to 7 s): $S_1 = \frac{1}{2}\times 2\times 10 = 10\text{ m}$
$\frac{S_1}{S} = \frac{10}{40} = \frac{1}{4}$ . So the answer is (b).

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Q85MHT CET 2024 · 9 MAY · SHIFT 1Easy

Graphs of Motion · Velocity-time graphs: distance & displacement

A velocity - time graph of a body is shown below. The distance covered by the body from 6 second to 9 second is

Show answer and solution

Answer: (c)

Distance = area under the $v$ - $t$ graph from 6 s to 9 s.
Slope from 6 s to 10 s: $a = \frac{40 - 20}{4} = 5\text{ m/s}^2$ , so $v(9) = 20 + 5\times 3 = 35\text{ m/s}$
Area = trapezium $= \frac{1}{2}(20 + 35)\times 3 = 82.5\text{ m}$
So the answer is (c) 82.5 m.

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Q86MHT CET 2024 · 2 MAY · SHIFT 1Easy

Graphs of Motion · Velocity-time graphs: slope, a-t graphs

The acceleration of a moving body can be found from

Show answer and solution

Answer: (c)

$a = \frac{dv}{dt}$ , which is the slope of the velocity-time graph.
Area under $v$ - $t$ gives displacement; slope of $x$ - $t$ gives velocity.
So the answer is (c).

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Q87NEET 2024Easy

Graphs of Motion · Velocity-time graphs: slope, a-t graphs

The velocity $(v)-$ time $(t)$ plot of the motion of a body is shown below:

The acceleration $(a)-$ time $(t)$ graph that best suits this motion is:

Show answer and solution

Answer: (c)

$a$ = slope of the $v$ - $t$ graph.
At rest at first: $a = 0$ . Rising straight line: constant positive $a$ . Flat part: $a = 0$ .
Falling straight line: constant negative $a$ . Then $v = 0$ : $a = 0$ .
So $a$ - $t$ is zero, a positive block, zero, a negative block. So the answer is (c).

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Q88JEE Main 2023 · 10 APR · SHIFT 1Easy

Graphs of Motion · Position-time graphs

The position-time graphs for two students A and B returning from the school to their homes are shown in figure.

(A) A lives closer to the school

(B) B lives closer to the school

(C) A takes lesser time to reach home

(D) A travels faster than B

(E) B travels faster than A

Choose the correct answer from the options given below:

Show answer and solution

Answer: (b)

Final (home) position of A is lower on the $x$ axis than that of B, so A lives closer: (A) true, (B) false.
A starts at $t = 0$ while B starts later and moves faster, so A takes more time: (C) false.
B's line is steeper, so B has the larger velocity: (E) true, (D) false.
So the answer is (b) (A) and (E) only.

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Q89JEE Main 2023 · 11 APR · SHIFT 1Easy

Graphs of Motion · Velocity-time graphs: distance & displacement

From the $v - t$ graph shown, the ratio of distance to displacement in $25\text{ s}$ of motion is:

Show answer and solution

Answer: (d)

Areas: 0-5 s: $25$ ; 5-10 s: $50$ ; 10-15 s: $\frac{1}{2}(10+20)\times 5 = 75$ ; 15-20 s: $\frac{1}{2}\times 5\times 20 = 50$ (all positive).
20-25 s: $\frac{1}{2}\times 5\times 20 = 50$ below the axis (negative).
Distance $= 25 + 50 + 75 + 50 + 50 = 250\text{ m}$ ; displacement $= 200 - 50 = 150\text{ m}$
Ratio $= \frac{250}{150} = \frac{5}{3}$ . So the answer is (d).

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Q90JEE Main 2023 · 24 JAN · SHIFT 2Easy

Graphs of Motion · Velocity-time graphs: distance & displacement

The velocity time graph of a body moving in a straight line is shown in the figure.

The ratio of displacement to distance travelled by the body in time 0 to 10s is:

Show answer and solution

Answer: (b)

Areas: 0-2 s: $8\times 2 = +16$ ; 2-4 s: $-4\times 2 = -8$ ; 4-8 s: $4\times 4 = +16$ ; 8-10 s: $-4\times 2 = -8$ (in m).
Displacement $= 16 - 8 + 16 - 8 = 16\text{ m}$
Distance $= 16 + 8 + 16 + 8 = 48\text{ m}$
Ratio $= 16 : 48 = 1 : 3$ . So the answer is (b).

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Q91JEE Main 2023 · 8 APR · SHIFT 2Easy

Graphs of Motion · Velocity-time graphs: slope, a-t graphs

Given below are two statements

Statement I: Area under velocity- time graph gives the distance travelled by the body in a given time. Statement II: Area under acceleration- time graph is equal to the change in velocity in the given time.

In the light of given statements, choose the correct answer from the options given below.

Show answer and solution

Answer: (c)

Statement I: area under $v$ - $t$ gives displacement, not distance (negative areas cancel). False.
Statement II: $\int a\,dt = \Delta v$ , so area under $a$ - $t$ is the change in velocity. True.
So the answer is (c).

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Q92JEE Main 2023 · 30 JAN · SHIFT 1Medium

Graphs of Motion · Identifying and matching graphs

Match Column-I with Column-II:

Choose the correct answer from the options given below:

Show answer and solution

Answer: (b)

$v$ = slope of the $x$ - $t$ graph.
A: parabola from origin, slope grows linearly from zero, so $v \propto t$ : II.
B: falls from $x_0$ with slope negative and shrinking to zero: IV.
C: triangle, constant $+v$ then constant $-v$ : III. D: straight line, constant $v$ : I.
A-II, B-IV, C-III, D-I. So the answer is (b).

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Q93NEET 2022 · PHASE 2Easy

Graphs of Motion · Position-time graphs

The position-time (x - t) graph for positive acceleration is

Show answer and solution

Answer: (b)

Slope of $x$ - $t$ graph is velocity; positive acceleration means slope increases with time.
So the graph must be concave upward.
Graph (b) curves upward with increasing slope.
So the answer is (b).

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Q94NEET 2022 · PHASE 1Easy

Graphs of Motion · Position-time graphs

The displacement-time graphs of two moving particles make angles of 30 $^\circ$ and 45 $^\circ$ with the x-axis as shown in the figure. The ratio of their respective velocity is

Show answer and solution

Answer: (d)

Velocity = slope of displacement-time graph $= \tan\theta$
$\dfrac{v_1}{v_2} = \dfrac{\tan 30^\circ}{\tan 45^\circ} = \dfrac{1/\sqrt{3}}{1}$
So the ratio is (d) $1 : \sqrt{3}$ .

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Q95JEE Main 2021 · 24 FEB · SHIFT 1Easy

Graphs of Motion · Velocity-time graphs: slope, a-t graphs

If the velocity-time graph has the shape AMB, what would be the shape of the corresponding acceleration-time graph?

Show answer and solution

Answer: (c)

$a$ = slope of the $v$ - $t$ graph.
A to M: straight line with negative slope, so $a$ is a negative constant.
M to B: straight line with positive slope, so $a$ is a positive constant.
So $a$ - $t$ is a negative step that jumps to a positive step at M. So the answer is (c).

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Q96JEE Main 2021 · 18 MAR · SHIFT 1Easy

Graphs of Motion · Identifying and matching graphs

The position, velocity and acceleration of a particle moving with a constant acceleration can be represented by:

Show answer and solution

Answer: (b)

Constant $a$ : $a$ - $t$ is a horizontal line.
$v = u + at$ : $v$ - $t$ is a straight line with slope $a$ .
$x = x_0 + ut + \frac{1}{2}at^2$ : $x$ - $t$ is a parabola curving upward.
Only option (b) has all three. So the answer is (b).

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Q97JEE Main 2021 · 31 AUG · SHIFT 2Easy

Graphs of Motion · Velocity-displacement graphs

A particle is moving with constant acceleration 'a'. Following graph shows v2 versus x(displacement) plot. The acceleration of the particle is ___________ m/s2.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 1

$v^2 = u^2 + 2ax$ , so slope of the $v^2$ - $x$ graph $= 2a$ .
Slope $= \frac{80 - 20}{30 - 0} = 2$
$2a = 2 \Rightarrow a = 1$
$a = 1\text{ m/s}^2$

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Q98JEE Main 2021 · 18 MAR · SHIFT 2Medium

Graphs of Motion · Velocity-displacement graphs

The velocity $-$ displacement graph of a particle is shown in the figure.

The acceleration $-$ displacement graph of the same particle is represented by:

Show answer and solution

Answer: (a)

From the graph $v = v_0\left(1 - \frac{x}{x_0}\right)$ , so $\frac{dv}{dx} = -\frac{v_0}{x_0}$ .
$a = v\frac{dv}{dx} = -\frac{v_0^2}{x_0}\left(1 - \frac{x}{x_0}\right)$
At $x = 0$ : $a = -\frac{v_0^2}{x_0}$ ; at $x = x_0$ : $a = 0$ ; linear in between.
Straight line from a negative value up to zero at $x_0$ . So the answer is (a).

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Q99JEE Main 2021 · 16 MAR · SHIFT 1Medium

Graphs of Motion · Velocity-displacement graphs

The velocity-displacement graph describing the motion of bicycle is shown in the figure.

The acceleration-displacement graph of the bicycle's motion is best described by:

Show answer and solution

Answer: (a)

For $0 \le x \le 200\text{ m}$ : $v = 10 + \frac{x}{5}$ , so $\frac{dv}{dx} = 0.2\text{ s}^{-1}$ .
$a = v\frac{dv}{dx} = 0.2\left(10 + \frac{x}{5}\right) = 2 + \frac{x}{25}$ : linear, starting at $2\text{ m/s}^2$ .
For $200 < x \le 400\text{ m}$ : $v$ = constant, so $a = 0$ .
Linear rise then zero matches graph (a). So the answer is (a).

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Q100JEE Main 2020 · 4 SEP · SHIFT 2Easy

Graphs of Motion · Velocity-time graphs: distance & displacement

The speed versus time graph for a particle is shown in the figure. The distance travelled (in m) by the particle during the time interval t = 0 to t = 5 s will be________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 20

Distance = area under the speed-time graph.
The line passes through the origin and reaches $8\text{ m/s}$ at $t = 5\text{ s}$ .
Area $= \frac{1}{2}\times 5\times 8 = 20$
Distance $= 20\text{ m}$

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Q101JEE Main 2020 · 5 SEP · SHIFT 2Medium

Graphs of Motion · Velocity-time graphs: distance & displacement

The velocity (v) and time (t) graph of a body in a straight line motion is shown in the figure. The point S is at 4.333 seconds. The total distance covered by the body in 6 s is:

Show answer and solution

Answer: (d)

Distance = sum of magnitudes of areas under the $v$ - $t$ graph.
0-2 s: $\frac{1}{2}\times 2\times 4 = 4$ ; 2-3 s: $1\times 4 = 4$ ; 3 s to $\frac{13}{3}$ s: $\frac{1}{2}\times\frac{4}{3}\times 4 = \frac{8}{3}$
$\frac{13}{3}$ s to 6 s (below axis): $\frac{1}{2}\times\frac{5}{3}\times 2 = \frac{5}{3}$
Total $= 8 + \frac{8}{3} + \frac{5}{3} = \frac{37}{3}\text{ m}$ . So the answer is (d).

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Q102JEE Main 2019 · 10 JAN · SHIFT 2Medium

Graphs of Motion · Velocity-time graphs: distance & displacement

A particle starts from the origin at time t = 0 and moves along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5s?

Show answer and solution

Answer: (b)

Starting at the origin, position = area under the $v$ - $t$ graph.
0-2 s: $\frac{1}{2}\times 2\times 2 = 2$ ; 2-4 s: $2\times 2 = 4$ ; 4-5 s: $3\times 1 = 3$
$x(5) = 2 + 4 + 3 = 9\text{ m}$ . So the answer is (b).

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Q103JEE Main 2019 · 8 APR · SHIFT 2Medium

Graphs of Motion · Identifying and matching graphs

A particle starts from origin O from rest and moves with a uniform acceleration along the positive x- axis. Identify all figures that correctly represent the motion qualitatively. (a = acceleration, v = velocity, x = displacement, t = time)

Show answer and solution

Answer: (d)

Uniform acceleration from rest at origin: $a$ = constant, $v = at$ , $x = \frac{1}{2}at^2$ .
(A) constant $a$ : correct. (B) $v$ - $t$ straight line through origin: correct.
(C) $x$ starts above zero and flattens out: wrong. (D) parabola from origin curving upward: correct.
So (A), (B), (D). So the answer is (d).

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Q104JEE Main 2018 · 15 APR · SHIFT 1Medium

Graphs of Motion · Velocity-time graphs: distance & displacement

The velocity-time graphs of a car and a scooter are shown in the figure. (i) The difference between the distance travelled by the car and the scooter in $15\text{ s}$ and (ii) the time at which the car will catch up with the scooter are, respectively.

Show answer and solution

Answer: (a)

In 15 s: car $= \frac{1}{2}\times 15\times 45 = 337.5\text{ m}$ , scooter $= 30\times 15 = 450\text{ m}$
Difference $= 450 - 337.5 = 112.5\text{ m}$
After 15 s the car (45 m/s) gains $15\text{ m/s}$ on the scooter: extra time $= \frac{112.5}{15} = 7.5\text{ s}$
Catch-up at $t = 15 + 7.5 = 22.5\text{ s}$ . So the answer is (a).

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Q105JEE Main 2018Medium

Graphs of Motion · Identifying and matching graphs

All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

Show answer and solution

Answer: (c)

The graphs describe a body thrown up with constant downward acceleration: $v$ falls linearly through zero, $v$ - $x$ is a parabola, $x$ - $t$ is an inverted parabola.
Distance never decreases and the speed is greatest at the start, so the distance-time graph must start with a steep slope.
Graph (c) starts with zero slope (speed zero at start), which is inconsistent.
So the answer is (c).

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Q106JEE Main 2017 · 8 APR · SHIFT 1Easy

Graphs of Motion · Velocity-time graphs: slope, a-t graphs

Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity?

Show answer and solution

Answer: (c)

With constant retardation $a$ : $v^2 = u^2 - 2ax$ , so $v = \sqrt{u^2 - 2ax}$ .
$v$ stays positive and falls slowly at first, then steeply, reaching zero at $x = \frac{u^2}{2a}$ with a vertical tangent.
This matches the velocity-distance curve that bends down to the $x$ -axis. So the answer is (c).

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Q107JEE Main 2008 · AIEEEMedium

Graphs of Motion · Identifying and matching graphs

A body is at rest at $x = 0.$ At $t = 0,$ it starts moving in the positive $x$ -direction with a constant acceleration. At the same instant another body passes through $x = 0$ moving in the positive $x$ direction with a constant speed. The position of the first body is given by $x_1(t)$ after time $\text{'}t\text{'};$ and that of the second body by $x_2(t)$ after the same time interval. Which of the following graphs correctly describes $(x_1 - x_2)$ as a function of time $\text{'}t\text{'}$ ?

Show answer and solution

Answer: (b)

$x_1 = \frac{1}{2}at^2$ , $x_2 = ut$ , so $x_1 - x_2 = \frac{1}{2}at^2 - ut$ .
This is an upward parabola through the origin: negative for $0 < t < \frac{2u}{a}$ , zero at $t = \frac{2u}{a}$ , then positive and rising.
So the answer is (b).

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Q108NEET 1994 · AIPMTEasy

Graphs of Motion · Position-time graphs

The displacement-time graph of a moving particle is shown below. The instantaneous velocity of the particle is negative at the point

Show answer and solution

Answer: (a)

Instantaneous velocity = slope of the tangent to the $x$ - $t$ graph.
At C and F the curve is rising (positive slope); at D the slope is zero (maximum).
At E the curve is falling, so the slope is negative.
So the answer is (a) E.

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Q109NEET 1992 · AIPMTEasy

Graphs of Motion · Identifying and matching graphs

Which of the following curve does not represent motion in one dimension?

Show answer and solution

Answer: (b)

In real motion a body has only one velocity at any instant.
The circle gives two values of $v$ at the same $t$ , and also needs time to run backward.
So it cannot represent motion. So the answer is (b).

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Q110MHT CET 2026 · 11 APR · SHIFT 1Medium

Equations of Kinematics · Stage motion & midpoint velocity

A car passes three points A, B and C at $3$ m/s, $6$ m/s and $9$ m/s respectively in a straight line with uniform acceleration. If distance $AB = 60$ m then find the distance BC.

Show answer and solution

Answer: (a)

Use $v^2 - u^2 = 2as$ on each part (same $a$ ).
AB: $6^2 - 3^2 = 27 = 2a(60)$
BC: $9^2 - 6^2 = 45 = 2a(BC)$
Dividing: $\frac{BC}{60} = \frac{45}{27} \Rightarrow BC = 100\text{ m}$
So the answer is (a) $100$ m.

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Q111MHT CET 2026 · 20 MAY · SHIFT 1Medium

Equations of Kinematics · Distance in the nth second

A body moving with uniform acceleration travel $16$ m in the third second of its motion and $24$ m in the fifth second of its motion. The velocity of the body in $10$ second is

Show answer and solution

Answer: (b)

$S_n = u + \frac{a}{2}(2n - 1)$ , so $S_5 - S_3 = 2a = 24 - 16 = 8 \Rightarrow a = 4\text{ m/s}^2$ .
$S_3 = u + \frac{4}{2}(5) = u + 10 = 16 \Rightarrow u = 6\text{ m/s}$
$v = u + at = 6 + 4(10) = 46\text{ m/s}$
So the answer is (b) $46$ m/s.

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Q112MHT CET 2025 · 20 APR · SHIFT 2Medium

Equations of Kinematics · Two bodies: meeting & chasing

Two bodies A and B move in same straight line starting from same position. Body A moves with constant velocity ' u ' and body B moves with constant acceleration ' $a$ '. When their velocities become equal, the distance between them is

Show answer and solution

Answer: (d)

Velocities are equal when $at = u \Rightarrow t = \frac{u}{a}$ .
$s_A = ut = \frac{u^2}{a}$ , $s_B = \frac{1}{2}at^2 = \frac{u^2}{2a}$
Gap $= \frac{u^2}{a} - \frac{u^2}{2a} = \frac{u^2}{2a}$
So the answer is (d) $\frac{u^2}{2a}$ .

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Q113MHT CET 2025 · 26 APR · SHIFT 2Medium

Equations of Kinematics · Stopping distance & penetration

A car moving at a speed ' v ' is stopped in a certain distance when the brakes produce a deceleration ' $a$ '. If the speed of the car is ' $nv$ ', what must be the deceleration of the car to stop it in the same distance and in the same time?

Show answer and solution

Answer: (c)

Stopping distance $s = \frac{v^2}{2a}$ .
Same distance with speed $nv$ : $\frac{(nv)^2}{2a'} = \frac{v^2}{2a} \Rightarrow a' = n^2a$ .
So the answer is (c) $n^2 \cdot a$ .

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Q114MHT CET 2025 · 21 APR · SHIFT 2Medium

Equations of Kinematics · Distance in the nth second

A body starts from rest and moves with a uniform acceleration. The ratio of the distance covered by the body in the $n$ th second of its motion to the total distance travelled in n second is

Show answer and solution

Answer: (b)

From rest: $S_n = \frac{a}{2}(2n - 1)$ and $S = \frac{1}{2}an^2$ .
$\frac{S_n}{S} = \frac{2n - 1}{n^2} = \frac{2}{n} - \frac{1}{n^2}$
So the answer is (b) $\frac{2}{n} - \frac{1}{n^2}$ .

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Q115MHT CET 2024 · 16 MAY · SHIFT 2Easy

Equations of Kinematics · Stage motion & midpoint velocity

A body travelling with uniform acceleration crosses two points A and B with velocities $20\text{ m/s}$ and $30\text{ m/s}$ respectively. The speed of the body at mid point of A and B is (nearly)

Show answer and solution

Answer: (b)

For uniform acceleration, speed at the midpoint: $v_m = \sqrt{\frac{v_A^2 + v_B^2}{2}}$ (from $v^2 = u^2 + 2as$ over each half).
$v_m = \sqrt{\frac{400 + 900}{2}} = \sqrt{650}$
$v_m \approx 25.5\text{ m/s}$ . So the answer is (b).

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Q116JEE Main 2024 · 9 APR · SHIFT 2Easy

Equations of Kinematics · Two bodies: meeting & chasing

Two cars are travelling towards each other at speed of $20\text{ m s}^{-1}$ each. When the cars are $300\text{ m}$ apart, both the drivers apply brakes and the cars retard at the rate of $2\text{ m s}^{-2}$ . The distance between them when they come to rest is:

Show answer and solution

Answer: (b)

Stopping distance of each car: $s = \frac{u^2}{2a} = \frac{20^2}{2 \times 2} = 100\text{ m}$ .
Both cars together cover $2 \times 100 = 200\text{ m}$ .
Gap left $= 300 - 200 = 100\text{ m}$ .
So the answer is (b) 100 m.

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Q117JEE Main 2024 · 4 APR · SHIFT 2Easy

Equations of Kinematics · Stopping distance & penetration

A bus moving along a straight highway with speed of $72\text{ km/h}$ is brought to a halt within $4\text{ s}$ after applying the brakes. The distance travelled by the bus during this time (Assume the retardation is uniform) is ________ $\text{m}$ .

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 40

$u = 72\text{ km/h} = 20\text{ m/s}$ , $v = 0$ , $t = 4\text{ s}$ .
For uniform retardation, $s = \frac{u + v}{2}t = \frac{20 + 0}{2} \times 4$
$s = 40\text{ m}$

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Q118JEE Main 2024 · 4 APR · SHIFT 1Easy

Equations of Kinematics · Distance in the nth second

A body travels $102.5\text{ m}$ in $n$ th second and $115.0\text{ m}$ in $(n+2)$ th second. The acceleration is:

Show answer and solution

Answer: (a)

$S_n = u + \frac{a}{2}(2n - 1)$ , so $S_{n+2} - S_n = \frac{a}{2}(4) = 2a$ .
$2a = 115.0 - 102.5 = 12.5$
$a = 6.25\text{ m/s}^2$
So the answer is (a) $6.25\text{ m/s}^2$ .

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Q119JEE Main 2024 · 29 JAN · SHIFT 1Medium

Equations of Kinematics · Equations of motion: direct use

A body starts moving from rest with constant acceleration covers displacement $S_1$ in first $(p-1)$ seconds and $S_2$ in first $p$ seconds. The displacement $S_1 + S_2$ will be made in time:

Show answer and solution

Answer: (d)

From rest, $S = \frac{1}{2}at^2$ : $S_1 = \frac{1}{2}a(p - 1)^2$ , $S_2 = \frac{1}{2}ap^2$ .
$S_1 + S_2 = \frac{1}{2}a\left[(p - 1)^2 + p^2\right] = \frac{1}{2}a(2p^2 - 2p + 1)$
Set equal to $\frac{1}{2}aT^2$ : $T = \sqrt{2p^2 - 2p + 1}\text{ s}$
So the answer is (d) $\sqrt{2p^2 - 2p + 1}$ s.

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Q120JEE Main 2024 · 30 JAN · SHIFT 1Medium

Equations of Kinematics · Stage motion & midpoint velocity

The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t+1)\text{ s}$ are $125\text{ m}$ and $50\text{ m/s}$ , respectively. The distance travelled by the particle in $(t+2)^{\text{th}}\text{ s}$ is _________ m.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 175

Increase in velocity in 1 s is 50 m/s, so $a = 50\text{ m/s}^2$ .
Displacement in $t$ to $t+1$ : $125 = v_t(1) + \frac{1}{2}(50)(1)^2 \Rightarrow v_t = 100\text{ m/s}$ .
The $(t+2)$ th second is from $t+1$ to $t+2$ , starting at $v_{t+1} = 150\text{ m/s}$ .
$s = 150(1) + \frac{1}{2}(50)(1)^2 = 175\text{ m}$

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Q121MHT CET 2024 · 11 MAY · SHIFT 1Medium

Equations of Kinematics · Stopping distance & penetration

A bullet is fired on a target with velocity V. Its velocity decreases from V to $V/2$ . When it penetrates 30 cm in a target. Through what thickness it will penetrate further in the target before coming to rest?

Show answer and solution

Answer: (c)

Constant retardation: $v^2 = u^2 - 2as$ , so the drop in $v^2$ is proportional to distance.
First 30 cm: $V^2 - \frac{V^2}{4} = \frac{3V^2}{4} = 2a(30)$
Rest of the way: $\frac{V^2}{4} = 2a(x)$
Dividing: $\frac{x}{30} = \frac{1}{3} \Rightarrow x = 10\text{ cm}$
So the answer is (c) 10 cm.

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Q122MHT CET 2024 · 2 MAY · SHIFT 2Medium

Equations of Kinematics · Stopping distance & penetration

A driver applies the brakes on seeing the red traffic signal 400 m ahead. At the time of applying brakes, the vehicle was moving with $15\text{ m/s}$ and retarding at $0.3\text{ m/s}^2$ . The distance of the vehicle from the traffic light one minute after application of brakes is

Show answer and solution

Answer: (d)

Time to stop $= \frac{15}{0.3} = 50\text{ s}$ , less than 60 s, so the vehicle is at rest after 1 minute.
Stopping distance $= \frac{u^2}{2a} = \frac{15^2}{2 \times 0.3} = 375\text{ m}$
Distance from the light $= 400 - 375 = 25\text{ m}$
So the answer is (d) 25 m.

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Q123JEE Main 2024 · 5 APR · SHIFT 1Medium

Equations of Kinematics · Distance in the nth second

A body moves on a frictionless plane starting from rest. If $S_n$ is distance moved between $t = n - 1$ and $t = n$ and $S_{n-1}$ is distance moved between $t = n - 2$ and $t = n - 1$ , then the ratio $\frac{S_{n-1}}{S_n}$ is $\left(1 - \frac{2}{x}\right)$ for $n = 10$ . The value of $x$ is __________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 19

From rest, $S_n = \frac{a}{2}(2n - 1)$ and $S_{n-1} = \frac{a}{2}(2n - 3)$ .
$\frac{S_{n-1}}{S_n} = \frac{2n - 3}{2n - 1} = 1 - \frac{2}{2n - 1}$
For $n = 10$ : $x = 2n - 1 = 19$

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Q124JEE Main 2023 · 6 APR · SHIFT 2Easy

Equations of Kinematics · Equations of motion: direct use

A particle starts with an initial velocity of $10.0\text{ ms}^{-1}$ along $x$ -direction and accelerates uniformly at the rate of $2.0\text{ ms}^{-2}$ . The time taken by the particle to reach the velocity of $60.0\text{ ms}^{-1}$ is __________.

Show answer and solution

Answer: (d)

$v = u + at$
$60 = 10 + 2t$
$t = 25\text{ s}$ . So the answer is (d).

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Q125MHT CET 2023 · 13 MAY · SHIFT 2Easy

Equations of Kinematics · Two bodies: meeting & chasing

Two bodies $A$ and $B$ start from the same point at the same instant and move along a straight line. Body $A$ moves with uniform acceleration $a$ and body $B$ moves with uniform velocity $v$ . They meet after time $t$ . The value of $t$ is

Show answer and solution

Answer: (a)

They meet when both displacements are equal: $s_A = \frac{1}{2}at^2$ , $s_B = vt$ .
$\frac{1}{2}at^2 = vt \Rightarrow t = \frac{2v}{a}$
So the answer is (a) $\frac{2v}{a}$ .

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Q126MHT CET 2023 · 11 MAY · SHIFT 1Medium

Equations of Kinematics · Stopping distance & penetration

A bullet is fired on a target with velocity ' $V$ '. Its velocity decreases from ' $V$ ' to ' $V/2$ ' when it penetrates $30\text{ cm}$ in a target. Through what thickness it will penetrate further in the target before coming to rest?

Show answer and solution

Answer: (c)

Constant retardation: drop in $v^2$ is proportional to distance.
First 30 cm: $V^2 - \frac{V^2}{4} = \frac{3V^2}{4} = 2a(30)$
Rest of the way: $\frac{V^2}{4} = 2a(x)$
$\frac{x}{30} = \frac{1}{3} \Rightarrow x = 10\text{ cm}$
So the answer is (c) 10 cm.

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Q127JEE Main 2023 · 1 FEB · SHIFT 2Medium

Equations of Kinematics · Stopping distance & penetration

For a train engine moving with speed of $20\text{ ms}^{-1}$ , the driver must apply brakes at a distance of 500 $\text{m}$ before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed $\sqrt{x}\text{ ms}^{-1}$ . The value of $x$ is ____________.

(Assuming same retardation is produced by brakes)

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 200

Retardation: $0 = 20^2 - 2a(500) \Rightarrow a = 0.4\text{ m/s}^2$ .
Braking over 250 m: $v^2 = 20^2 - 2(0.4)(250) = 400 - 200 = 200$
$v = \sqrt{200}\text{ m/s}$ , so $x = 200$

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Q128JEE Main 2022 · 29 JUN · SHIFT 2Easy

Equations of Kinematics · Equations of motion: direct use

A small toy starts moving from the position of rest under a constant acceleration. If it travels a distance of 10m in t s, the distance travelled by the toy in the next t s will be:

Show answer and solution

Answer: (c)

From rest $s = \frac{1}{2}at^2$ , so $s \propto t^2$ .
In $2t$ : $s = 4\times 10 = 40\text{ m}$
Distance in the next $t$ s $= 40 - 10 = 30\text{ m}$ . So the answer is (c).

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Q129JEE Main 2022 · 30 JUL · SHIFT 2Easy

Equations of Kinematics · Two bodies: meeting & chasing

Two trucks A and B are approaching each other on a straight path with the velocities of $16\text{ m/s}$ and $20\text{ m/s}$ respectively. When they are $200\text{ m}$ apart their drivers see each other and applies the breaks simultaneously. If the truck A decelerates with $2\text{ m/s}^2$ and truck B decelerates with $4\text{ m/s}^2$ , what is the distance between them when they finally stop?

Show answer and solution

Answer: (c)

Stopping distance $s = \frac{u^2}{2a}$ for each truck.
Truck A: $s_A = \frac{16^2}{2 \times 2} = 64\text{ m}$ ; truck B: $s_B = \frac{20^2}{2 \times 4} = 50\text{ m}$ .
Gap left $= 200 - (64 + 50) = 86\text{ m}$ .
So the answer is (c) $86\text{ m}$ .

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Q130JEE Main 2022 · 25 JUL · SHIFT 1Easy

Equations of Kinematics · Stopping distance & penetration

A car is moving with speed of $150\text{ km/h}$ and after applying the brake it will move $27\text{ m}$ before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling ___________ m distance.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 3

For the same retardation, $s \propto u^2$ .
Speed becomes $\frac{1}{3}$ , so distance becomes $\frac{1}{9}$ .
$s' = \frac{27}{9} = 3\text{ m}$

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Q131JEE Main 2022 · 27 JUL · SHIFT 2Medium

Equations of Kinematics · Stopping distance & penetration

The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at (4 + x) cm inside the block. The value of x is:

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Answer: (c)

Constant resistance means constant retardation, so the drop in $v^2$ is proportional to distance.
First 4 cm: $u^2 - \frac{u^2}{9} = \frac{8u^2}{9} = 2a(4)$
Further $x$ : $\frac{u^2}{9} = 2a(x)$
$\frac{x}{4} = \frac{1}{8} \Rightarrow x = 0.5\text{ cm}$
So the answer is (c) 0.5.

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Q132JEE Main 2021 · 26 FEB · SHIFT 2Easy

Equations of Kinematics · Stage motion & midpoint velocity

A scooter accelerates from rest for time t1 at constant rate a1 and then retards at constant rate a2 for time t2 and comes to rest. The correct value of $\frac{t_1}{t_2}$ will be:

Show answer and solution

Answer: (c)

Peak speed reached after accelerating: $v = a_1t_1$
Retarding to rest: $0 = v - a_2t_2 \Rightarrow v = a_2t_2$
$a_1t_1 = a_2t_2 \Rightarrow \frac{t_1}{t_2} = \frac{a_2}{a_1}$ . So the answer is (c).

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Q133JEE Main 2021 · 25 FEB · SHIFT 1Easy

Equations of Kinematics · Stage motion & midpoint velocity

An engine of a train, moving with uniform acceleration, passes the signal-post with velocity u and the last compartment with velocity v. The velocity with which middle point of the train passes the signal post is:

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Answer: (d)

Let train length be $L$ : $v^2 = u^2 + 2aL$ .
Middle point passes after the train moves $\frac{L}{2}$ : $v_m^2 = u^2 + 2a\frac{L}{2} = u^2 + \frac{v^2 - u^2}{2}$
$v_m = \sqrt{\frac{u^2 + v^2}{2}}$ . So the answer is (d).

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Q134MHT CET 2021 · 20 SEP · SHIFT 1Easy

Equations of Kinematics · Two bodies: meeting & chasing

Two bodies ' $A$ ' and ' $B$ ' start from the same point at the same instant and move along a straight line. 'A' moves with uniform acceleration (a) and 'B' moves with uniform velocity (V). They meet after time 't'. The value of 't' is

Show answer and solution

Answer: (a)

They meet when both displacements are equal: $s_A = \frac{1}{2}at^2$ , $s_B = Vt$ .
$\frac{1}{2}at^2 = Vt \Rightarrow t = \frac{2V}{a}$
So the answer is (a) $\frac{2V}{a}$ .

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Q135JEE Main 2021 · 17 MAR · SHIFT 1Medium

Equations of Kinematics · Stage motion & midpoint velocity

A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If the total time elapsed is t seconds, the total distance travelled is:

Show answer and solution

Answer: (c)

Maximum speed: $\frac{v}{\alpha} + \frac{v}{\beta} = t \Rightarrow v = \frac{\alpha\beta t}{\alpha + \beta}$ .
Distance = area of the triangular $v$ - $t$ graph $= \frac{1}{2}vt$ .
$s = \frac{1}{2} \cdot \frac{\alpha\beta t}{\alpha + \beta} \cdot t = \frac{\alpha\beta}{2(\alpha + \beta)}t^2$
So the answer is (c) $\frac{\alpha\beta}{2(\alpha+\beta)}t^2$ .

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Q136MHT CET 2021 · 22 SEP · SHIFT 2Medium

Equations of Kinematics · Stopping distance & penetration

A driver applies the brakes on seeing the red traffic signal $400\text{ m}$ ahead. At the time of applying the brakes, the vehicle was moving with $15\text{ m/s}$ and retarding at $0.3\text{ m/s}^2$ . The distance covered by the vehicle from the traffic light 1 minute after the application of brakes is

Show answer and solution

Answer: (a)

Time to stop $= \frac{15}{0.3} = 50\text{ s} < 60\text{ s}$ , so the vehicle has stopped.
Stopping distance $= \frac{15^2}{2 \times 0.3} = 375\text{ m}$
Distance from the light $= 400 - 375 = 25\text{ m}$
So the answer is (a) 25 m.

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Q137NEET 2021Medium

Equations of Kinematics · Distance in the nth second

A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let Sn be the distance travelled by the block in the interval t = n $-$ 1 to t = n. Then the ratio $\frac{S_n}{S_{n+1}}$ is:

Show answer and solution

Answer: (c)

Smooth incline: constant acceleration from rest, so $S_n = \frac{a}{2}(2n - 1)$ .
$S_{n+1} = \frac{a}{2}(2n + 1)$
$\frac{S_n}{S_{n+1}} = \frac{2n - 1}{2n + 1}$
So the answer is (c) $\frac{2n-1}{2n+1}$ .

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Q138MHT CET 2020 · 16 OCT · SHIFT 1Easy

Equations of Kinematics · Equations of motion: direct use

A moving body is covering distances which are proportional to square of the time. Then, the acceleration of the body is

Show answer and solution

Answer: (c)

$s \propto t^2$ means $s = kt^2$ .
$v = \frac{ds}{dt} = 2kt$ , $a = \frac{dv}{dt} = 2k$
So acceleration is constant and non-zero. So the answer is (c).

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Q139JEE Main 2019 · 11 JAN · SHIFT 2Medium

Equations of Kinematics · Equations of motion: direct use

A particle moves from the point $\left(2.0\hat{i} + 4.0\hat{j}\right)$ m, at t = 0, with an initial velocity $\left(5.0\hat{i} + 4.0\hat{j}\right)$ ms ${}^{-1}$ . It is acted upon by a constant force which produces a constant acceleration $\left(4.0\hat{i} + 4.0\hat{j}\right)$ ms ${}^{-2}$ . What is the distance of the particle from the origin at time 2 s?

Show answer and solution

Answer: (b)

$\vec{r} = \vec{r}_0 + \vec{u}t + \frac{1}{2}\vec{a}t^2$ with $t = 2\text{ s}$ .
x: $2 + 5(2) + \frac{1}{2}(4)(4) = 20\text{ m}$ ; y: $4 + 4(2) + \frac{1}{2}(4)(4) = 20\text{ m}$
Distance from origin $= \sqrt{20^2 + 20^2} = 20\sqrt{2}\text{ m}$
So the answer is (b) $20\sqrt{2}$ m.

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Q140JEE Main 2019 · 9 JAN · SHIFT 2Difficult

Equations of Kinematics · Stage motion & midpoint velocity

In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed ' $v$ ' more than that of car B. Both the cars start from rest and travel with constant acceleration a1 and a2 respectively. Then ' $v$ ' is equal to:

Show answer and solution

Answer: (c)

Same track length $d$ , from rest: $t_i = \sqrt{\frac{2d}{a_i}}$ , $v_i = \sqrt{2da_i}$ .
$v = v_1 - v_2 = \sqrt{2d}(\sqrt{a_1} - \sqrt{a_2})$
$t = t_2 - t_1 = \sqrt{2d}\,\frac{\sqrt{a_1} - \sqrt{a_2}}{\sqrt{a_1a_2}}$
$\frac{v}{t} = \sqrt{a_1a_2} \Rightarrow v = \sqrt{a_1a_2}\,t$
So the answer is (c) $\sqrt{a_1a_2}\,t$ .

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Q141JEE Main 2018 · 15 APR · SHIFT 1Easy

Equations of Kinematics · Stopping distance & penetration

An automobile, travelling at $40$ km/h, can be stopped at a distance of $40$ m by applying brakes. If the same automobile is travelling at $80$ km/h, the minimum stopping distance, in metres, is (assume no skidding):

Show answer and solution

Answer: (d)

For the same retardation, $s = \frac{u^2}{2a} \Rightarrow s \propto u^2$ .
Speed doubles, so $s' = 4 \times 40 = 160\text{ m}$ .
So the answer is (d) $160$ m.

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Q142JEE Main 2017 · 9 APR · SHIFT 1Easy

Equations of Kinematics · Two bodies: meeting & chasing

A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2. The car will catch up with the bus after a time of:

Show answer and solution

Answer: (c)

Relative to the bus, the car starts from rest with $a_{rel} = 4 - 2 = 2\text{ m/s}^2$ and must cover 200 m.
$200 = \frac{1}{2}(2)t^2 \Rightarrow t^2 = 200$
$t = 10\sqrt{2}\text{ s}$
So the answer is (c) $10\sqrt{2}\text{ s}$ .

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Q143NEET 2009 · AIPMTEasy

Equations of Kinematics · Equations of motion: direct use

A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is S1 and that covered in the first 20 seconds is S2, then

Show answer and solution

Answer: (b)

Constant force means constant $a$ ; from rest $S = \frac{1}{2}at^2$ , so $S \propto t^2$ .
$\frac{S_2}{S_1} = \left(\frac{20}{10}\right)^2 = 4$
So the answer is (b) $S_2 = 4S_1$ .

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Q144NEET 2008 · AIPMTEasy

Equations of Kinematics · Equations of motion: direct use

A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms ${}^{-1}$ to 20 ms ${}^{-1}$ while passing through a distance 135 m in t second. The value of t is

Show answer and solution

Answer: (b)

For uniform acceleration $s = \frac{u + v}{2}\,t$ .
$135 = \frac{10 + 20}{2}\,t = 15t$
$t = 9\text{ s}$ . So the answer is (b).

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Q145NEET 2008 · AIPMTEasy

Equations of Kinematics · Distance in the nth second

The distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3}$ m s ${}^{-2}$ , in the third second is

Show answer and solution

Answer: (a)

Distance in the $n$ th second: $S_n = u + \frac{a}{2}(2n - 1)$ , with $u = 0$ .
$S_3 = \frac{1}{2} \times \frac{4}{3} \times (2 \times 3 - 1) = \frac{2}{3} \times 5$
$S_3 = \frac{10}{3}\text{ m}$
So the answer is (a) $\frac{10}{3}$ m.

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Q146JEE Main 2005 · AIEEEDifficult

Equations of Kinematics · Stage motion & midpoint velocity

A car starting from rest accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed is 15 S, then

Show answer and solution

Answer: (d)

After distance $S$ : $v^2 = 2fS$ .
Deceleration at $\frac{f}{2}$ : distance $= \frac{v^2}{2(f/2)} = \frac{2fS}{f} = 2S$ .
Constant-speed part: $vt = 15S - S - 2S = 12S \Rightarrow v = \frac{12S}{t}$
$\frac{144S^2}{t^2} = 2fS \Rightarrow S = \frac{1}{72}ft^2$
So the answer is (d) $S = \frac{1}{72}ft^2$ .

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Q147JEE Main 2004 · AIEEEEasy

Equations of Kinematics · Stopping distance & penetration

An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 m. If the car is going twice as fast, i.e 120 km/h, the stopping distance will be

Show answer and solution

Answer: (d)

For the same retardation, $s \propto u^2$ .
Speed doubles, so $s' = 4 \times 20 = 80\text{ m}$ .
So the answer is (d) 80 m.

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Q148JEE Main 2003 · AIEEEEasy

Equations of Kinematics · Stopping distance & penetration

A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is

Show answer and solution

Answer: (c)

For the same retardation, $s \propto u^2$ .
Speed doubles (50 to 100 km/h), so $s' = 4 \times 6 = 24\text{ m}$ .
So the answer is (c) 24 m.

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Q149JEE Main 2002 · AIEEEEasy

Equations of Kinematics · Stopping distance & penetration

Speeds of two identical cars are $u$ and $4u$ at the specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is:

Show answer and solution

Answer: (d)

Identical cars have the same braking retardation, so $s = \frac{u^2}{2a} \Rightarrow s \propto u^2$ .
$\frac{s_1}{s_2} = \frac{u^2}{(4u)^2} = \frac{1}{16}$
So the answer is (d) $1 : 16$ .

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Q150NEET 1998 · AIPMTEasy

Equations of Kinematics · Stopping distance & penetration

A car moving with a speed of $40\text{ km/h}$ can be stopped by applying brakes after atleast $2\text{ m}$ . If the same car is moving with a speed of $80\text{ km/h}$ , what is the minimum stopping distance?

Show answer and solution

Answer: (a)

For the same retardation, $s = \frac{u^2}{2a} \Rightarrow s \propto u^2$ .
Speed doubles (40 to 80 km/h), so distance becomes $2^2 = 4$ times.
$s' = 4 \times 2 = 8\text{ m}$
So the answer is (a) $8\text{ m}$ .

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Q151NEET 1997 · AIPMTEasy

Equations of Kinematics · Equations of motion: direct use

If a car at rest accelerates uniformly to a speed of 144 km/h in 20 sec, it covers a distance of

Show answer and solution

Answer: (d)

$v = 144\text{ km/h} = 144\times\frac{5}{18} = 40\text{ m/s}$
$s = \frac{u + v}{2}\,t = \frac{0 + 40}{2}\times 20$
$s = 400\text{ m}$ . So the answer is (d).

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Q152NEET 1994 · AIPMTEasy

Equations of Kinematics · Equations of motion: direct use

The velocity of train increases uniformly from 20 km/h to 60 km/h in 4 hours. The distance travelled by the train during this period is

Show answer and solution

Answer: (a)

For uniform acceleration, average velocity $= \frac{u + v}{2} = \frac{20 + 60}{2} = 40\text{ km/h}$
$s = 40\times 4 = 160\text{ km}$
So the answer is (a).

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Q153NEET 1994 · AIPMTMedium

Equations of Kinematics · Stage motion & midpoint velocity

A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ and comes to rest. If total time elapsed is $t$ , then maximum velocity acquired by car will be

Show answer and solution

Answer: (d)

Let the maximum speed be $v$ . Time to speed up $t_1 = \frac{v}{\alpha}$ , time to stop $t_2 = \frac{v}{\beta}$ .
$t = \frac{v}{\alpha} + \frac{v}{\beta} = v\frac{\alpha + \beta}{\alpha\beta}$
$v = \frac{\alpha\beta t}{\alpha + \beta}$
So the answer is (d) $\frac{\alpha\beta t}{\alpha + \beta}$ .

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Q154NEET 1993 · AIPMTEasy

Equations of Kinematics · Distance in the nth second

A body starts from rest, what is the ratio of the distance travelled by the body during the 4th and 3rd second?

Show answer and solution

Answer: (a)

From rest, $S_n = \frac{a}{2}(2n - 1)$ .
$\frac{S_4}{S_3} = \frac{2(4) - 1}{2(3) - 1} = \frac{7}{5}$
So the answer is (a) $\frac{7}{5}$ .

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Q155NEET 1988 · AIPMTEasy

Equations of Kinematics · Stage motion & midpoint velocity

A car is moving along a straight road with a uniform acceleration. It passes through two points $P$ and $Q$ separated by a distance with velocity 30 km/h and 40 km/h respectively. The velocity of the car midway between $P$ and $Q$ is

Show answer and solution

Answer: (c)

Midpoint velocity: $v_m = \sqrt{\frac{v_P^2 + v_Q^2}{2}}$
$v_m = \sqrt{\frac{30^2 + 40^2}{2}} = \sqrt{1250}$
$v_m = 25\sqrt{2}\text{ km/h}$ . So the answer is (c).

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Q156NEET 2026Easy

Motion Under Gravity · Free fall: h, t and v

When a ruler falls vertically, 5 different persons catch it with different reaction times. ( $g = 9.8\text{ m s}^{-2}$ )

A. Person A has reaction time of 0.20 s.

B. Person $B$ has reaction time of 0.22 s.

C. Person C has reaction time of 0.18 s.

D. Person D has reaction time of 0.19 s.

E. Person E has reaction time of 0.21 s.

What is the correct order of the distance travelled by the ruler for each person?

Show answer and solution

Answer: (c)

The ruler falls freely during the reaction time: $d = \frac{1}{2}gt^2$ , so $d \propto t^2$ .
Longer reaction time means longer fall: $t_B (0.22) > t_E (0.21) > t_A (0.20) > t_D (0.19) > t_C (0.18)$ .
Order: B > E > A > D > C
So the answer is (c).

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Q157JEE Main 2026 · 5 APR · SHIFT 1Easy

Motion Under Gravity · Free fall: h, t and v

From 18 m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is ______ m.

(Take $g = 10\text{ m/s}^2$ and neglect the air resistance)

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 13

Need $v = g = 10$ m/s (numerically).
Distance fallen: $s = \frac{v^2}{2g} = \frac{100}{20} = 5$ m
Height above ground $= 18 - 5 = 13$ m
So the answer is 13 m.

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Q158MHT CET 2026 · 15 APR · SHIFT 1Easy

Motion Under Gravity · Two bodies under gravity

A stone falls from the top of a tower of height $100$ m and at the same time another stone is projected vertically upwards from the ground with a velocity $25$ m/s. The two stones meet after

Show answer and solution

Answer: (c)

Both stones have the same acceleration $g$ , so relative acceleration is zero.
Relative speed $= 25$ m/s, relative distance $= 100$ m
$t = \frac{100}{25} = 4$ s
So the answer is (c) 4 s.

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Q159NEET 2026Easy

Motion Under Gravity · Bounce & graphs under gravity

The following plots show variation of velocity $(v)$ with time $(t)$ of a ball thrown vertically upward, and falling back. Which of the following plots is/are correct?

Show answer and solution

Answer: (a)

Taking upward as positive, $v = u - gt$ : a straight line with constant negative slope $-g$ .
$v$ starts positive, becomes zero at the top, then becomes negative while falling, so the line crosses the $t$ -axis and continues below it: graph C.
A and E show speed (never negative), B rises then falls, D starts negative with positive slope, so none of these is correct.
So the answer is (a) C only.

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Q160JEE Main 2026 · 23 JAN · SHIFT 2Medium

Motion Under Gravity · Free fall: h, t and v

A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts decelerating with $3\text{ m/s}^2$ . At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is $5\text{ m/s}$ . The initial height of the airplane is ______ m.

$(g = 10\text{ m/s}^2)$

Show answer and solution

Answer: (a)

Free fall for 2 s: $v = 20$ m/s, $h_1 = \frac{1}{2} \times 10 \times 2^2 = 20$ m
Decelerating at 3 m/s $^2$ from 20 m/s to 5 m/s: $h_2 = \frac{20^2 - 5^2}{2 \times 3} = \frac{375}{6} = 62.5$ m
Initial height $= 20 + 62.5 + 10 = 92.5$ m
So the answer is (a) 92.5 m.

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Q161MHT CET 2026 · 19 MAY · SHIFT 1Medium

Motion Under Gravity · Drops & balls at regular intervals

A juggler throws balls into air. He throws one ball whenever the previous one is at its highest point. If he throws 4 balls in one second, the height from which each ball returns is nearly [acceleration due to gravity is $9.8$ m/s ${}^{2}$ ]

Show answer and solution

Answer: (a)

4 balls per second: interval $= 0.25$ s, which equals the time to reach the top.
$H = \frac{1}{2}gt^2 = \frac{1}{2} \times 9.8 \times 0.25^2 \approx 0.306$ m
$H \approx 30$ cm. So the answer is (a) 30 cm.

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Q162JEE Main 2026 · 28 JAN · SHIFT 1Medium

Motion Under Gravity · Drops & balls at regular intervals

Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from ground, at the instant when the first drop strikes the ground is ______ m.

$(g = 10\text{ m/s}^2)$

Show answer and solution

Answer: (c)

Time to fall 5 m: $T = \sqrt{2 \times 5/10} = 1$ s
Drop 1 lands as drop 6 starts: 5 intervals in 1 s, so interval $= 0.2$ s.
Drop 4 has fallen for $2 \times 0.2 = 0.4$ s: $s = \frac{1}{2} \times 10 \times 0.4^2 = 0.8$ m
Height above ground $= 5 - 0.8 = 4.2$ m. So the answer is (c) 4.2 m.

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Q163JEE Main 2026 · 8 APR · SHIFT 2Medium

Motion Under Gravity · Balloon & helicopter

A gas balloon is going up with a constant velocity of $10\text{ m/s}$ . When this balloon reached a height of 75 m, a stone is dropped from it and balloon keeps moving up with the same velocity. The height of the balloon when the stone hits the ground is ______ m. (Take $g = 10\text{ m/s}^2$ )

Show answer and solution

Answer: (d)

Stone leaves with the balloon's velocity, $10\text{ m/s}$ upward, at height 75 m.
$-75 = 10t - 5t^2 \Rightarrow t^2 - 2t - 15 = 0 \Rightarrow t = 5\text{ s}$
Balloon height $= 75 + 10 \times 5 = 125\text{ m}$
So the answer is (d) 125.

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Q164MHT CET 2026 · 16 APR · SHIFT 1Medium

Motion Under Gravity · Bounce & graphs under gravity

A ball is released from height 'h' which makes perfectly elastic collision with ground. The frequency of periodic vibratory motion is (g=acceleration due to gravity)

Show answer and solution

Answer: (a)

Time to fall height $h$ : $t = \sqrt{\frac{2h}{g}}$
Elastic bounce: it rises back to $h$ in the same time, so period $T = 2\sqrt{\frac{2h}{g}}$
$f = \frac{1}{T} = \frac{1}{2}\sqrt{\frac{g}{2h}}$
So the answer is (a) $\frac{1}{2}\sqrt{\frac{g}{2h}}$ .

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Q165MHT CET 2025 · 26 APR · SHIFT 1Easy

Motion Under Gravity · Thrown vertically: up, down, tower

A boy throws a ball vertically upwards from a bridge with velocity $5\text{ m/s}$ . It strikes water surface after 2 s. The height of the bridge is (Take $g = 10\text{ m/s}^2$ )

Show answer and solution

Answer: (d)

Take upward as positive: $s = ut - \frac{1}{2}gt^2$
$s = 5 \times 2 - \frac{1}{2} \times 10 \times 2^2 = 10 - 20 = -10$ m
The water is 10 m below the bridge. So the answer is (d) 10 m.

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Q166MHT CET 2024 · 15 MAY · SHIFT 2Easy

Motion Under Gravity · Free fall: h, t and v

A ball is released from the top of a tower of height Hm. It takes T second to reach the ground. The height of the ball from the ground after $\frac{T}{4}$ second is

Show answer and solution

Answer: (b)

Free fall from rest: $h = \frac{1}{2}gt^2 \Rightarrow h \propto t^2$ , with $H = \frac{1}{2}gT^2$ .
Fallen in $\frac{T}{4}$ : $\frac{H}{16}$
Height above ground $= H - \frac{H}{16} = \frac{15H}{16}$
So the answer is (b) $\frac{15H}{16}$ .

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Q167JEE Main 2024 · 27 JAN · SHIFT 2Medium

Motion Under Gravity · Free fall: h, t and v

A body falling under gravity covers two points $A$ and $B$ separated by $80\text{ m}$ in $2\text{ s}$ . The distance of upper point A from the starting point is _________ $\text{m}$ (use $g = 10\text{ ms}^{-2}$ ).

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 45

From A to B: $80 = v_A(2) + \frac{1}{2}(10)(2)^2 = 2v_A + 20 \Rightarrow v_A = 30$ m/s
Distance of A from start: $s = \frac{v_A^2}{2g} = \frac{900}{20} = 45$ m
So the answer is 45 m.

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Q168JEE Main 2024 · 6 APR · SHIFT 2Medium

Motion Under Gravity · Thrown vertically: up, down, tower

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in $t_1$ . If it is projected vertically downwards from the same point with the same speed, it reaches the ground in $t_2$ . Time required to reach the ground, if it is dropped from the top of the tower, is:

Show answer and solution

Answer: (b)

Up throw: $-h = ut_1 - \frac{1}{2}gt_1^2$ ; down throw: $h = ut_2 + \frac{1}{2}gt_2^2$
Eliminate $u$ (multiply first by $t_2$ , second by $t_1$ , subtract): $h(t_1 + t_2) = \frac{1}{2}g t_1 t_2 (t_1 + t_2) \Rightarrow h = \frac{1}{2}gt_1t_2$
Dropped: $h = \frac{1}{2}gt^2 \Rightarrow t = \sqrt{t_1t_2}$
So the answer is (b) $\sqrt{t_1t_2}$ .

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Q169MHT CET 2023 · 10 MAY · SHIFT 1Easy

Motion Under Gravity · Free fall: h, t and v

A body is released from the top of a tower ' $H$ ' metre high. It takes $t$ second to reach the ground. The height of the body $\frac{t}{2}$ second after release is

Show answer and solution

Answer: (c)

Free fall from rest: $h \propto t^2$ .
Fallen in $\frac{t}{2}$ : $\frac{H}{4}$
Height above ground $= H - \frac{H}{4} = \frac{3H}{4}$
So the answer is (c) $\frac{3H}{4}$ metre from ground.

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Q170JEE Main 2023 · 12 APR · SHIFT 1Easy

Motion Under Gravity · Thrown vertically: up, down, tower

A ball is thrown vertically upward with an initial velocity of $150\text{ m/s}$ . The ratio of velocity after $3\text{ s}$ and $5\text{ s}$ is $\frac{x+1}{x}$ . The value of $x$ is ___________.

$\{$ take, $g = 10\text{ m/s}^2\}$

Show answer and solution

Answer: (c)

$v = u - gt$ : $v_3 = 150 - 30 = 120$ m/s, $v_5 = 150 - 50 = 100$ m/s
$\frac{v_3}{v_5} = \frac{120}{100} = \frac{6}{5} = \frac{x+1}{x}$
$\Rightarrow x = 5$ . So the answer is (c) 5.

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Q171MHT CET 2023 · 13 MAY · SHIFT 2Medium

Motion Under Gravity · Free fall: h, t and v

A small steel ball is dropped from a height of $1.5\text{ m}$ into a glycerine jar. The ball reaches the bottom of the jar $1.5\text{ s}$ after it was dropped. If the retardation is $2.66\text{ m/s}^2$ , the height of the glycerine in the jar is about (acceleration due to gravity $g = 9.8\text{ m/s}^2$ )

Show answer and solution

Answer: (c)

Speed on reaching glycerine: $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1.5} \approx 5.42$ m/s, in time $t_1 = v/g \approx 0.55$ s
Read literally, time in glycerine $= 1.5 - 0.55 = 0.95$ s: $s = 5.42 \times 0.95 - \frac{1}{2} \times 2.66 \times 0.95^2 \approx 3.9$ m
The intended reading takes 1.5 s as the time inside the glycerine: $s = 5.42 \times 1.5 - \frac{1}{2} \times 2.66 \times 1.5^2 \approx 8.13 - 2.99 \approx 5.1$ m
This is nearest to 5.5 m. So the answer is (c) 5.5 m.

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Q172MHT CET 2023 · 12 MAY · SHIFT 2Medium

Motion Under Gravity · Thrown vertically: up, down, tower

A ball is projected vertically upwards from ground. It reaches a height ' $h$ ' in time $t_1$ , continues its motion and then takes a time $t_2$ to reach ground. The height $h$ in terms of $g, t_1$ and $t_2$ is $(g =$ acceleration due to gravity)

Show answer and solution

Answer: (b)

$h = ut - \frac{1}{2}gt^2$ has two roots: going up at $t_1$ , coming down at $t_1'$ .
Total flight time is $t_1 + t_2$ ; by symmetry the ball is at $h$ again at $t_1' = (t_1 + t_2) - t_1 = t_2$ .
Product of roots of $\frac{1}{2}gt^2 - ut + h = 0$ : $t_1 t_2 = \frac{2h}{g}$
$h = \frac{1}{2}gt_1t_2$ . So the answer is (b).

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Q173JEE Main 2023 · 29 JAN · SHIFT 1Medium

Motion Under Gravity · Bounce & graphs under gravity

A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height 5.0 m. Ball comes in contact with the floor for 0.2s. The average acceleration during contact is ___________ ms $-2$ .

(Given g = 10 ms $-2$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 120

Speed on hitting: $v_1 = \sqrt{2gh_1} = \sqrt{2(10)(9.8)} = 14\text{ m/s}$ (down)
Speed on rebound: $v_2 = \sqrt{2(10)(5)} = 10\text{ m/s}$ (up)
$a = \frac{\Delta v}{\Delta t} = \frac{10 - (-14)}{0.2} = \frac{24}{0.2}$
$a = 120\text{ m s}^{-2}$

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Q174NEET 2022 · PHASE 1Easy

Motion Under Gravity · Free fall: h, t and v

The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second

Show answer and solution

Answer: (c)

From rest, $S_n = \frac{g}{2}(2n - 1)$ .
$n = 1, 2, 3, 4$ gives $1 : 3 : 5 : 7$ (Galileo's law of odd numbers).
So the answer is (c) 1: 3: 5: 7.

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Q175JEE Main 2022 · 28 JUL · SHIFT 1Easy

Motion Under Gravity · Free fall: h, t and v

An NCC parade is going at a uniform speed of $9\text{ km/h}$ under a mango tree on which a monkey is sitting at a height of $19.6\text{ m}$ . At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is: (Given $g = 9.8\text{ m/s}^2$ )

Show answer and solution

Answer: (a)

Time of fall: $t = \sqrt{2h/g} = \sqrt{2 \times 19.6/9.8} = 2$ s
Parade speed: $9$ km/h $= 9 \times \frac{5}{18} = 2.5$ m/s
The cadet must be $d = 2.5 \times 2 = 5$ m from the tree at the instant of drop.
So the answer is (a) 5 m.

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Q176JEE Main 2022 · 26 JUN · SHIFT 1Easy

Motion Under Gravity · Free fall: h, t and v

A ball of mass 0.5 kg is dropped from the height of 10 m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ________ m. [Use g = 10 m/s2]

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 5

Need $v = g = 10$ m/s (numerically); mass does not matter.
Distance fallen: $s = \frac{v^2}{2g} = \frac{100}{20} = 5$ m
Height above ground $= 10 - 5 = 5$ m
So the answer is 5 m.

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Q177JEE Main 2022 · 24 JUN · SHIFT 1Easy

Motion Under Gravity · Thrown vertically: up, down, tower

From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in ____________ s.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 3

For a tower, time to fall from rest $t = \sqrt{t_1 t_2}$ , where $t_1, t_2$ are the up-throw and down-throw times with the same speed.
$t = \sqrt{6 \times 1.5} = \sqrt{9} = 3$ s
So the answer is 3 s.

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Q178JEE Main 2022 · 29 JUL · SHIFT 2Medium

Motion Under Gravity · Free fall: h, t and v

A ball is released from a height h. If $t_1$ and $t_2$ be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between $t_1$ and $t_2$ .

Show answer and solution

Answer: (d)

First half: $\frac{h}{2} = \frac{1}{2}gt_1^2 \Rightarrow t_1 = \sqrt{h/g}$
Whole height: $T = \sqrt{2h/g} = \sqrt{2}\,t_1$
$t_2 = T - t_1 = (\sqrt{2} - 1)t_1$
So the answer is (d) $t_2 = (\sqrt{2} - 1)t_1$ .

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Q179JEE Main 2022 · 29 JUL · SHIFT 2Medium

Motion Under Gravity · Drops & balls at regular intervals

A juggler throws balls vertically upwards with same initial velocity in air. When the first ball reaches its highest position, he throws the next ball. Assuming the juggler throws n balls per second, the maximum height the balls can reach is

Show answer and solution

Answer: (d)

$n$ balls per second: interval $= \frac{1}{n}$ s, which equals the time to reach the top.
$H = \frac{1}{2}gt^2 = \frac{1}{2}g\left(\frac{1}{n}\right)^2$
So the answer is (d) $\frac{g}{2n^2}$ .

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Q180JEE Main 2022 · 29 JUL · SHIFT 1Medium

Motion Under Gravity · Thrown vertically: up, down, tower

A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height $\frac{h}{3}$ while going up and coming down respectively.

Show answer and solution

Answer: (b)

Maximum height $h$ gives $u^2 = 2gh$ .
At height $\frac{h}{3}$ : $\frac{h}{3} = ut - \frac{1}{2}gt^2 \Rightarrow t = \frac{u \pm \sqrt{u^2 - \frac{2gh}{3}}}{g}$
$u = \sqrt{2gh}$ and $\sqrt{u^2 - \frac{2gh}{3}} = \sqrt{\frac{4gh}{3}}$
$\frac{t_1}{t_2} = \frac{\sqrt{2} - \sqrt{4/3}}{\sqrt{2} + \sqrt{4/3}} = \frac{\sqrt{6} - 2}{\sqrt{6} + 2} = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$
So the answer is (b) $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ .

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Q181JEE Main 2022 · 28 JUL · SHIFT 2Medium

Motion Under Gravity · Thrown vertically: up, down, tower

A ball is thrown vertically upwards with a velocity of $19.6\text{ ms}^{-1}$ from the top of a tower. The ball strikes the ground after $6\text{ s}$ . The height from the ground up to which the ball can rise will be $\left(\frac{k}{5}\right)\text{ m}$ . The value of $k$ is __________. (use $g = 9.8\text{ m/s}^2$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 392

Upward positive, origin at top of tower: $-H = ut - \frac{1}{2}gt^2 = 19.6(6) - 4.9(36)$
$-H = 117.6 - 176.4 = -58.8 \Rightarrow H = 58.8\text{ m}$
Rise above the tower: $\frac{u^2}{2g} = \frac{19.6^2}{19.6} = 19.6\text{ m}$
Maximum height above ground $= 58.8 + 19.6 = 78.4\text{ m} = \frac{k}{5}$
$k = 392$

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Q182JEE Main 2022 · 29 JUN · SHIFT 1Medium

Motion Under Gravity · Two bodies under gravity

Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms ${}^{-1}$ . [use g = 10 ms ${}^{-2}$ ]:

Show answer and solution

Answer: (d)

Meeting point is $180 - 100 = 80\text{ m}$ below the top.
Ball A: $80 = \frac{1}{2}(10)t^2 \Rightarrow t = 4\text{ s}$
Ball B moves for $4 - 2 = 2\text{ s}$ : $80 = 2u + \frac{1}{2}(10)(2)^2 = 2u + 20$
$u = 30\text{ m/s}$ , so the answer is (d) 30.

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Q183JEE Main 2022 · 26 JUN · SHIFT 2Medium

Motion Under Gravity · Two bodies under gravity

A ball is projected vertically upward with an initial velocity of 50 ms ${}^{-1}$ at t = 0s. At t = 2s, another ball is projected vertically upward with same velocity. At t = __________ s, second ball will meet the first ball (g = 10 ms ${}^{-2}$ ).

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 6

$y_1 = 50t - 5t^2$ , $y_2 = 50(t-2) - 5(t-2)^2$
Meet when $y_1 = y_2$ : $0 = -100 + 20t - 20$
$t = 6\text{ s}$

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Q184JEE Main 2022 · 27 JUL · SHIFT 1Medium

Motion Under Gravity · Bounce & graphs under gravity

A bullet is shot vertically downwards with an initial velocity of $100\text{ m/s}$ from a certain height. Within 10 s, the bullet reaches the ground and instantaneously comes to rest due to the perfectly inelastic collision. The velocity-time curve for total time $t = 20\text{ s}$ will be:

(Take g = 10 m/s2).

Show answer and solution

Answer: (a)

Take upward as positive: $v = -100 - 10t$
At $t = 10\text{ s}$ : $v = -200\text{ m/s}$ , a straight line from $-100$ to $-200$ .
Perfectly inelastic hit: velocity drops to zero at once and stays zero up to 20 s.
So the answer is (a).

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Q185MHT CET 2021 · 21 SEP · SHIFT 2Easy

Motion Under Gravity · Free fall: h, t and v

A body at rest falls through a height 'h' with velocity 'V'. If it has to fall down further for its velocity to become three times, the distance travelled in that interval is

Show answer and solution

Answer: (a)

From rest, $v^2 = 2gh \Rightarrow h \propto v^2$ .
Speed $3V$ needs total fall $9h$ .
Further distance $= 9h - h = 8h$
So the answer is (a) 8 h.

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Q186MHT CET 2021 · 23 SEP · SHIFT 1Easy

Motion Under Gravity · Drops & balls at regular intervals

A student is throwing balls vertically upwards such that he throws the $2$ nd ball when the $1$ st ball reaches maximum height. If he throws balls at an interval of 3 second, the maximum height of the balls is $\left(g = 10\,\frac{\text{m}}{\text{s}^2}\right)$

Show answer and solution

Answer: (a)

The next ball is thrown when the previous one is at the top, so time to reach the top $= 3$ s.
$u = gt = 10 \times 3 = 30$ m/s
$H = \frac{u^2}{2g} = \frac{900}{20} = 45$ m
So the answer is (a) 45 m.

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Q187JEE Main 2021 · 27 AUG · SHIFT 2Easy

Motion Under Gravity · Drops & balls at regular intervals

Water drops are falling from a nozzle of a shower onto the floor, from a height of 9.8 m. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor.

Show answer and solution

Answer: (d)

Time to fall 9.8 m: $T = \sqrt{2 \times 9.8/9.8} = \sqrt{2}$ s
Two intervals in $T$ , so drop 2 has fallen for $t = \frac{\sqrt{2}}{2}$ s, i.e. $t^2 = 0.5$ s $^2$ .
Distance fallen $= \frac{1}{2} \times 9.8 \times 0.5 = 2.45$ m
Height above floor $= 9.8 - 2.45 = 7.35$ m. So the answer is (d) 7.35 m.

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Q188JEE Main 2021 · 25 FEB · SHIFT 2Medium

Motion Under Gravity · Free fall: h, t and v

A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another stone starts to fall from a point 25 m below the top. Both stones reach the bottom of building simultaneously. The height of the building is:

Show answer and solution

Answer: (c)

Stone 1 after falling 5 m: $v = \sqrt{2 \times 10 \times 5} = 10$ m/s (take $g = 10$ m/s $^2$ ).
For the remaining time $t$ : stone 1 covers $H - 5 = 10t + 5t^2$ ; stone 2 covers $H - 25 = 5t^2$
Subtract: $20 = 10t \Rightarrow t = 2$ s
$H = 25 + 5(2)^2 = 45$ m. So the answer is (c) 45 m.

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Q189JEE Main 2021 · 25 JUL · SHIFT 1Medium

Motion Under Gravity · Drops & balls at regular intervals

Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at 4th second after its fall to the next droplet is 34.3 m. At what rate the droplets are coming from the tap? (Take g = 9.8 m/s2)

Show answer and solution

Answer: (c)

Droplet at 4 s has fallen $\frac{1}{2} \times 9.8 \times 4^2 = 78.4$ m.
Next droplet (released $T$ s later) has fallen $78.4 - 34.3 = 44.1$ m: $44.1 = 4.9(4 - T)^2$
$(4 - T)^2 = 9 \Rightarrow 4 - T = 3 \Rightarrow T = 1$ s
So the answer is (c) 1 drop / second.

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Q190JEE Main 2021 · 27 JUL · SHIFT 1Medium

Motion Under Gravity · Thrown vertically: up, down, tower

A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching $\frac{h}{3}$ in both the directions.

Show answer and solution

Answer: (c)

$u^2 = 2gh$ . At height $\frac{h}{3}$ : $\frac{h}{3} = ut - \frac{1}{2}gt^2$
Roots: $t = \frac{\sqrt{2gh} \pm \sqrt{4gh/3}}{g}$
$\frac{t_1}{t_2} = \frac{\sqrt{2} - \sqrt{4/3}}{\sqrt{2} + \sqrt{4/3}}$ ; multiply top and bottom by $\sqrt{3/2}$ : $\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$
So the answer is (c) $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ .

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Q191JEE Main 2021 · 26 AUG · SHIFT 1Medium

Motion Under Gravity · Two bodies under gravity

Two spherical balls having equal masses with radius of 5 cm each are thrown upwards along the same vertical direction at an interval of 3s with the same initial velocity of 35 m/s, then these balls collide at a height of ............... m. (Take g = 10 m/s2)

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 50

$y_1 = 35t - 5t^2$ , $y_2 = 35(t-3) - 5(t-3)^2$
Equal heights: $0 = -105 + 30t - 45 \Rightarrow t = 5\text{ s}$
Height $= 35(5) - 5(5)^2 = 175 - 125 = 50\text{ m}$ (the 5 cm radius is negligible).
Collision height $= 50\text{ m}$

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Q192JEE Main 2021 · 25 JUL · SHIFT 2Medium

Motion Under Gravity · Balloon & helicopter

A balloon was moving upwards with a uniform velocity of 10 m/s. An object of finite mass is dropped from the balloon when it was at a height of 75 m from the ground level. The height of the balloon from the ground when object strikes the ground was around:

(take the value of g as 10 m/s2)

Show answer and solution

Answer: (c)

Object starts with $10\text{ m/s}$ upward from 75 m.
$-75 = 10t - 5t^2 \Rightarrow t = 5\text{ s}$
Balloon height $= 75 + 10(5) = 125\text{ m}$
So the answer is (c) 125 m.

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Q193JEE Main 2021 · 17 MAR · SHIFT 2Difficult

Motion Under Gravity · Bounce & graphs under gravity

A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to $\frac{81}{100}$ of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms ${}^{-2}$ )

Show answer and solution

Answer: (a)

Height ratio $e^2 = 0.81 \Rightarrow$ speed ratio $e = 0.9$ per bounce.
Total distance $= h\frac{1 + e^2}{1 - e^2} = 5 \times \frac{1.81}{0.19} \approx 47.6\text{ m}$
First fall time $t_0 = \sqrt{\frac{2h}{g}} = 1\text{ s}$ ; total time $= t_0\frac{1 + e}{1 - e} = \frac{1.9}{0.1} = 19\text{ s}$
Average speed $= \frac{47.6}{19} \approx 2.5\text{ m/s}$
So the answer is (a) 2.50 m/s.

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Q194NEET 2020 · PHASE 1Easy

Motion Under Gravity · Thrown vertically: up, down, tower

A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is: (g = 10 m/s2)

Show answer and solution

Answer: (c)

$v^2 = u^2 + 2gh$
$h = \frac{80^2 - 20^2}{2 \times 10} = \frac{6400 - 400}{20} = 300$ m
So the answer is (c) 300 m.

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Q195NEET 2020 · PHASE 2Medium

Motion Under Gravity · Free fall: h, t and v

A person sitting in the ground floor of a building notices through the window, of height 1.5 m, a ball dropped from the roof of the building crosses the window in 0.1 s. What is the velocity of the ball when it is at the topmost point of the window? ( $g = 10\text{ m/s}^2$ )

Show answer and solution

Answer: (c)

Let $v$ be the speed at the top of the window: $s = vt + \frac{1}{2}gt^2$
$1.5 = v(0.1) + \frac{1}{2}(10)(0.1)^2 = 0.1v + 0.05$
$v = \frac{1.45}{0.1} = 14.5$ m/s
So the answer is (c) 14.5 m/s.

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Q196JEE Main 2020 · 8 JAN · SHIFT 2Medium

Motion Under Gravity · Free fall: h, t and v

A ball is dropped from the top of a 100 m high tower on a planet. In the last $\frac{1}{2}\text{ s}$ before hitting the ground, it covers a distance of 19 m. Acceleration due to gravity (in ms–2) near the surface on that planet is _____.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 8

Total time $T$ : $100 = \frac{1}{2}gT^2$ ; before the last 0.5 s it fell $100 - 19 = 81$ m: $81 = \frac{1}{2}g(T - 0.5)^2$
Divide: $\frac{T}{T - 0.5} = \sqrt{\frac{100}{81}} = \frac{10}{9} \Rightarrow T = 5$ s
$g = \frac{2 \times 100}{5^2} = 8$ m/s $^2$
So the answer is 8 m/s $^2$ .

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Q197JEE Main 2020 · 5 SEP · SHIFT 1Medium

Motion Under Gravity · Balloon & helicopter

A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height h. The time taken by the packet to reach the ground is close to:

[g is the acceleration due to gravity]

______

Show answer and solution

Answer: (a)

At height $h$ the helicopter's speed: $v^2 = 2gh \Rightarrow v = \sqrt{2gh}$ upward; the packet starts with this velocity.
$-h = \sqrt{2gh}\,t - \frac{1}{2}gt^2$ . Put $t = x\sqrt{h/g}$ : $x^2 - 2\sqrt{2}x - 2 = 0$
$x = \sqrt{2} + 2 \approx 3.41$
So $t \approx 3.4\sqrt{\frac{h}{g}}$ , option (a).

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Q198JEE Main 2020 · 4 SEP · SHIFT 1Medium

Motion Under Gravity · Bounce & graphs under gravity

A Tennis ball is released from a height h and after freely falling on a wooden floor it rebounds and reaches height $\frac{h}{2}$ . The velocity versus height of the ball during its motion may be represented graphically by:

(graph are drawn schematically and on not to scale)

Show answer and solution

Answer: (c)

Falling from $h$ (downward, $v < 0$ ): $v^2 = 2g(h - y)$ , so $|v|$ grows from 0 at $y = h$ to $\sqrt{2gh}$ at $y = 0$ along a curve.
At the floor the velocity reverses: it jumps to a positive value $\sqrt{gh}$ at $y = 0$ .
Rising: $v^2 = 2g(\frac{h}{2} - y)$ , so $v$ falls to 0 at $y = \frac{h}{2}$ .
Curve from $h$ to the negative $v$ -axis, then from the positive $v$ -axis to $\frac{h}{2}$ : answer (c).

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Q199NEET 2019Easy

Motion Under Gravity · Free fall: h, t and v

A person standing on the floor of an elevator drops a coin. The coin reaches the floor in time $t_1$ if the elevator is at rest and in time $t_2$ if the elevator is moving uniformly. Then

Show answer and solution

Answer: (a)

In a lift moving with uniform velocity, the frame is inertial, so the coin's acceleration relative to the floor is still $g$ .
The coin starts at rest relative to the floor and falls the same height $h$ , so $t = \sqrt{2h/g}$ in both cases.
So the answer is (a) $t_1 = t_2$ .

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Q200JEE Main 2017Easy

Motion Under Gravity · Bounce & graphs under gravity

A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?

Show answer and solution

Answer: (d)

Upward positive: $v = u - gt$ , a straight line of constant slope $-g$ .
It starts at $+u$ , passes through $v = 0$ at the highest point and continues to negative values on the way down, without any change of slope.
Graphs (a), (b) and (c) change slope or direction, which would need a change in acceleration.
So the answer is (d).

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Q201JEE Main 2015Difficult

Motion Under Gravity · Bounce & graphs under gravity

Two stones are thrown up simultaneously from the edge of a cliff $240\text{ m}$ high with initial speed of $10\text{ m/s}$ and $40\text{ m/s}$ respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first?

(Assume stones do not rebound after hitting the ground and neglect air resistance, take $g = 10\text{ m/s}^2$ )

(The figures are schematic and not drawn to scale)

Show answer and solution

Answer: (a)

Both have the same acceleration, so $y_2 - y_1 = (40 - 10)t = 30t$ (straight line) while both are in flight.
Stone 1 lands: $-240 = 10t - 5t^2 \Rightarrow t = 8\text{ s}$ ; then $y_2 - y_1 = 240\text{ m}$ .
After that $y_1 = -240$ is fixed: $y_2 - y_1 = 40t - 5t^2 + 240$ , a downward-curving parabola.
Stone 2 lands when $t^2 - 8t - 48 = 0 \Rightarrow t = 12\text{ s}$ , where $y_2 - y_1 = 0$ .
So the answer is (a).

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Q202JEE Main 2014Medium

Motion Under Gravity · Thrown vertically: up, down, tower

From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:

Show answer and solution

Answer: (c)

Time to reach the highest point: $t_0 = \frac{u}{g}$ , so time to hit the ground $t = \frac{nu}{g}$ .
Upward positive, origin at top of tower: $-H = ut - \frac{1}{2}gt^2$
$-H = \frac{nu^2}{g} - \frac{n^2u^2}{2g} \Rightarrow 2gH = n^2u^2 - 2nu^2$
So the answer is (c) $2gH = nu^2(n-2)$ .

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Q203NEET 2013Easy

Motion Under Gravity · Free fall: h, t and v

A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h1, h2 and h3 is

Show answer and solution

Answer: (d)

From rest, distances in equal successive time intervals are in the ratio $1:3:5$ (Galileo's law of odd numbers).
$h_1 = \frac{1}{2}g(5)^2 = 12.5g$ , $h_2 = \frac{1}{2}g(10^2-5^2) = 37.5g$ , $h_3 = \frac{1}{2}g(15^2-10^2) = 62.5g$
So $h_2 = 3h_1$ , $h_3 = 5h_1$ , i.e. $h_1 = \frac{h_2}{3} = \frac{h_3}{5}$ .
So the answer is (d).

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Q204NEET 2011 · AIPMT · PRELIMSEasy

Motion Under Gravity · Free fall: h, t and v

A boy standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 m s ${}^{-2}$ , the velocity with which it hits the ground is

Show answer and solution

Answer: (b)

Dropped from rest: $v = \sqrt{2gh}$
$v = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20$ m/s
So the answer is (b) 20.0 m/s.

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Q205NEET 2010 · AIPMT · PRELIMSMedium

Motion Under Gravity · Two bodies under gravity

A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v?

(Take g = 10 m/s2)

Show answer and solution

Answer: (a)

First ball at $t = 18\text{ s}$ has fallen $\frac{1}{2}(10)(18)^2 = 1620\text{ m}$ .
Second ball moves for $18 - 6 = 12\text{ s}$ : $12v + \frac{1}{2}(10)(12)^2 = 12v + 720$
They meet: $12v + 720 = 1620 \Rightarrow v = 75\text{ m/s}$
So the answer is (a) 75 m/s.

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Q206JEE Main 2009 · AIEEEMedium

Motion Under Gravity · Bounce & graphs under gravity

Consider a rubber ball freely falling from a height $h = 4.9\text{ m}$ onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as a function of time will be:

Show answer and solution

Answer: (b)

Upward positive: falling, $v$ goes linearly from 0 to $-v_1$ in time $t_1$ .
Elastic bounce: $v$ jumps from $-v_1$ to $+v_1$ at once.
Then $v$ falls linearly (slope $-g$ ) from $+v_1$ through 0 at $2t_1$ to $-v_1$ at $3t_1$ , and repeats.
Height: parabolic arcs, each returning to the same height $h$ .
So the answer is (b).

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Q207NEET 2006 · AIPMTEasy

Motion Under Gravity · Free fall: h, t and v

Two bodies A (of mass 1 kg) and B (of mass 3 kg) are dropped from heights of 16 m and 25 m, respectively. The ratio of the time taken by them to reach the ground is

Show answer and solution

Answer: (a)

Time of fall $t = \sqrt{2h/g}$ does not depend on mass, so $t \propto \sqrt{h}$ .
$\frac{t_A}{t_B} = \sqrt{\frac{16}{25}} = \frac{4}{5}$
So the answer is (a) 4/5.

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Q208NEET 2005 · AIPMTEasy

Motion Under Gravity · Thrown vertically: up, down, tower

A ball is thrown vertically upward. It has a speed of 10 m/sec when it has reached one half of its maximum height. How high does the ball rise?

(Take g = 10 m/s2.)

Show answer and solution

Answer: (a)

Maximum height: $u^2 = 2gH$
At $H/2$ : $v^2 = u^2 - 2g\frac{H}{2} = 2gH - gH = gH$
$10^2 = 10 \times H \Rightarrow H = 10$ m
So the answer is (a) 10 m.

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Q209JEE Main 2005 · AIEEEMedium

Motion Under Gravity · Free fall: h, t and v

A parachutist after bailing out falls $50\text{ m}$ without friction. When parachute opens, it decelerates at $2\text{ m/s}^2.$ He reaches the ground with a speed of $3\text{ m/s}$ . At what height, did he bail out?

Show answer and solution

Answer: (d)

Speed after 50 m free fall: $v^2 = 2gh = 2 \times 9.8 \times 50 = 980$
With deceleration 2 m/s $^2$ down to 3 m/s: $s = \frac{980 - 9}{2 \times 2} \approx 243$ m
Total height $= 50 + 243 = 293$ m
So the answer is (d) 293 m.

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Q210JEE Main 2004 · AIEEEEasy

Motion Under Gravity · Free fall: h, t and v

A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in $\frac{T}{3}$ seconds?

Show answer and solution

Answer: (a)

$h = \frac{1}{2}gT^2$ ; distance fallen in $T/3$ : $s = \frac{1}{2}g\left(\frac{T}{3}\right)^2 = \frac{h}{9}$
Height above ground $= h - \frac{h}{9} = \frac{8h}{9}$
So the answer is (a) $\frac{8h}{9}$ m from the ground.

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Q211NEET 2003 · AIPMTMedium

Motion Under Gravity · Drops & balls at regular intervals

A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time? (Given g = 9.8 m/s2)

Show answer and solution

Answer: (a)

For more than two balls in the air, each ball must stay up longer than two intervals: $T > 2 \times 2 = 4$ s
Time of flight $T = \frac{2u}{g} > 4 \Rightarrow u > 2g = 19.6$ m/s
So the answer is (a) more than 19.6 m/s.

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Q212NEET 2003 · AIPMTMedium

Motion Under Gravity · Thrown vertically: up, down, tower

If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is

Show answer and solution

Answer: (b)

By symmetry, the last $t$ seconds of ascent mirror the first $t$ seconds of a fall from rest at the top.
Distance $= \frac{1}{2}gt^2$ , independent of $u$ .
So the answer is (b) $\frac{1}{2}gt^2$ .

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Q213JEE Main 2002 · AIEEEEasy

Motion Under Gravity · Thrown vertically: up, down, tower

From a building two balls A and B are thrown such that A is thrown upwards and B downwards ( both vertically with the same speed ). If vA and vB are their respective velocities on reaching the ground, then

Show answer and solution

Answer: (b)

Ball A comes back to the throwing point with the same speed $u$ , now moving down, just like B.
Both then fall the same height $h$ : $v^2 = u^2 + 2gh$ for each.
So the answer is (b) $v_A = v_B$ .

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Q214NEET 2001 · AIPMTEasy

Motion Under Gravity · Thrown vertically: up, down, tower

A particle is thrown vertically upward. Its velocity at half of the height is 10 m/s, then the maximum height attained by it (g = 10 m/s2)

Show answer and solution

Answer: (c)

Maximum height: $u^2 = 2gH$ ; at $H/2$ : $v^2 = u^2 - gH = gH$
$10^2 = 10 \times H \Rightarrow H = 10$ m
So the answer is (c) 10 m.

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Q215NEET 1996 · AIPMTEasy

Motion Under Gravity · Free fall: h, t and v

A body dropped from a height $h$ with initial velocity zero, strikes the ground with a velocity 3 m/s. Another body of same mass dropped from the same height $h$ with an initial velocity of 4 m/s. The final velocity of second mass, with which it strikes the ground is

Show answer and solution

Answer: (a)

First body: $3^2 = 0 + 2gh \Rightarrow 2gh = 9$
Second body: $v^2 = u^2 + 2gh = 4^2 + 9 = 25$
$v = 5$ m/s. So the answer is (a) 5 m/s.

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Q216NEET 1995 · AIPMTEasy

Motion Under Gravity · Drops & balls at regular intervals

The water drop falls at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap at instant the first drop touches the ground. How far above the ground is the second drop at that instant?

Show answer and solution

Answer: (a)

Time for a drop to fall 5 m: $T = \sqrt{2 \times 5/10} = 1$ s
When drop 1 lands, drop 3 is just leaving, so two intervals fit in 1 s: interval $= 0.5$ s.
Drop 2 has fallen for $0.5$ s: $s = \frac{1}{2} \times 10 \times 0.5^2 = 1.25$ m
Height above ground $= 5 - 1.25 = 3.75$ m. So the answer is (a) 3.75 m.

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Q217NEET 1992 · AIPMTMedium

Motion Under Gravity · Free fall: h, t and v

A body dropped from top of a tower fall through 40 m during the last two seconds of its fall. The height of tower is ( $g = 10\text{ m/s}^2$ )

Show answer and solution

Answer: (b)

Total time $T$ : $\frac{1}{2}gT^2 - \frac{1}{2}g(T-2)^2 = 40$
$5[T^2 - (T-2)^2] = 5(4T - 4) = 40 \Rightarrow T = 3$ s
$H = \frac{1}{2} \times 10 \times 3^2 = 45$ m
So the answer is (b) 45 m.

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Q218NEET 1989 · AIPMTEasy

Motion Under Gravity · Free fall: h, t and v

What will be the ratio of the distance moved by a freely falling body from rest in 4th and 5th seconds of journey?

Show answer and solution

Answer: (b)

Distance in the $n$ th second from rest: $s_n = \frac{g}{2}(2n-1)$
$s_4 : s_5 = (2\cdot4-1) : (2\cdot5-1) = 7 : 9$
So the answer is (b) 7 : 9.

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Q219JEE Main 2026 · 4 APR · SHIFT 2Difficult

Relative Velocity · Relative velocity along a line

Two cars $A$ and $B$ are moving in the same direction along a straight line with speeds $100\text{ km/h}$ and $80\text{ km/h}$ , respectively such that car $A$ is moving ahead of car $B$ . A person in car $B$ throws a stone with a speed $v$ so that it hits the car $A$ with a speed of $5\text{ m/s}$ . The value of $v$ is $\underline{\qquad}\text{ km/h}$ .

Show answer and solution

Answer: (c)

Stone must hit A with $5\text{ m/s} = 18\text{ km/h}$ relative to A, so stone's ground speed $= 100 + 18 = 118\text{ km/h}$ .
Speed relative to B (thrower): $v = 118 - 80 = 38\text{ km/h}$
So the answer is (c) 38.

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Q220NEET 2025Difficult

Relative Velocity · Relative velocity along a line

Two cities $X$ and $Y$ are connected by a regular bus service with a bus leaving in either direction every $T$ min. A girl is driving scooty with a speed of $60\text{ km/h}$ in the direction $X$ to $Y$ notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period $T$ of the bus service and the speed (assumed constant) of the buses.

Show answer and solution

Answer: (b)

Bus speed $v$ ; buses in each direction are spaced $vT$ apart.
Same direction: $\frac{vT}{v - 60} = 30\text{ min}$ ; opposite: $\frac{vT}{v + 60} = 10\text{ min}$
Divide: $\frac{v + 60}{v - 60} = 3 \Rightarrow v = 120\text{ km/h}$
$\frac{120T}{60} = 30 \Rightarrow T = 15\text{ min}$
So the answer is (b) 15 min, 120 km/h.

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Q221JEE Main 2024 · 1 FEB · SHIFT 2Easy

Relative Velocity · Relative velocity along a line

Train A is moving along two parallel rail tracks towards north with speed $72\text{ km/h}$ and train B is moving towards south with speed $108\text{ km/h}$ . Velocity of train B with respect to A and velocity of ground with respect to B are (in $\text{ms}^{-1}$ ):

Show answer and solution

Answer: (b)

North positive: $v_A = 72 \times \frac{5}{18} = 20\text{ m/s}$ , $v_B = -108 \times \frac{5}{18} = -30\text{ m/s}$
$v_{BA} = v_B - v_A = -30 - 20 = -50\text{ m/s}$
$v_{GB} = 0 - v_B = +30\text{ m/s}$
So the answer is (b) $-50$ and $30$ .

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Q222MHT CET 2023 · 10 MAY · SHIFT 2Easy

Relative Velocity · Relative velocity along a line

Two trains, each $30\text{ m}$ long are travelling in opposite directions with velocities $5\text{ m/s}$ and $10\text{ m/s}$ . They will cross after

Show answer and solution

Answer: (a)

Opposite directions: relative speed $= 5 + 10 = 15\text{ m/s}$
Distance to cross $= 30 + 30 = 60\text{ m}$
$t = \frac{60}{15} = 4\text{ s}$ , so the answer is (a) 4 s.

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Q223JEE Main 2023 · 13 APR · SHIFT 2Easy

Relative Velocity · Relative velocity along a line

A passenger sitting in a train A moving at $90\text{ km/h}$ observes another train $B$ moving in the opposite direction for $8\text{ s}$ . If the velocity of the train B is $54\text{ km/h}$ , then length of train B is:

Show answer and solution

Answer: (d)

$v_A = 90\text{ km/h} = 25\text{ m/s}$ , $v_B = 54\text{ km/h} = 15\text{ m/s}$
Opposite directions: relative speed $= 40\text{ m/s}$
Length of B $= 40 \times 8 = 320\text{ m}$ , so the answer is (d).

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Q224JEE Main 2023 · 13 APR · SHIFT 1Medium

Relative Velocity · Relative velocity along a line

Two trains 'A' and 'B' of length ' $l$ ' and ' $4l$ ' are travelling into a tunnel of length ' $L$ ' in parallel tracks from opposite directions with velocities $108\text{ km/h}$ and $72\text{ km/h}$ , respectively. If train 'A' takes $35\text{ s}$ less time than train 'B' to cross the tunnel, then length ' $L$ ' of tunnel is:

(Given $L = 60l$ )

Show answer and solution

Answer: (c)

$v_A = 108\text{ km/h} = 30\text{ m/s}$ , $v_B = 72\text{ km/h} = 20\text{ m/s}$
Time to cross tunnel: $t_A = \frac{L + l}{30}$ , $t_B = \frac{L + 4l}{20}$
With $L = 60l$ : $\frac{64l}{20} - \frac{61l}{30} = 35 \Rightarrow \frac{7l}{6} = 35 \Rightarrow l = 30\text{ m}$
$L = 60 \times 30 = 1800\text{ m}$ , so the answer is (c).

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Q225JEE Main 2021 · 20 JUL · SHIFT 2Medium

Relative Velocity · Escalator problems

A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time t1. If he remains stationary on a moving escalator then the escalator takes him up in time t2. The time taken by him to walk up on the moving escalator will be:

Show answer and solution

Answer: (c)

Boy's speed $\frac{L}{t_1}$ , escalator speed $\frac{L}{t_2}$ ; speeds add.
$t = \frac{L}{L/t_1 + L/t_2} = \frac{t_1t_2}{t_1 + t_2}$
So the answer is (c).

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Q226JEE Main 2020 · 2 SEP · SHIFT 1Easy

Relative Velocity · Relative velocity along a line

Train A and train B are running on parallel tracks in the opposite directions with speeds of 36 km/hour and 72 km/hour, respectively. A person is walking in train A in the direction opposite to its motion with a speed of 1.8 km/ hour. Speed (in ms–1) of this person as observed from train B will be close to:

(take the distance between the tracks as negligible)

Show answer and solution

Answer: (b)

$v_A = 36\text{ km/h} = 10\text{ m/s}$ , person relative to A $= -1.8\text{ km/h} = -0.5\text{ m/s}$ , $v_B = -72\text{ km/h} = -20\text{ m/s}$
Person w.r.t. ground: $10 - 0.5 = 9.5\text{ m/s}$
Person w.r.t. B: $9.5 - (-20) = 29.5\text{ m/s}$
So the answer is (b) 29.5 m/s.

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Q227JEE Main 2019 · 12 JAN · SHIFT 1Medium

Relative Velocity · Relative velocity along a line

A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of 30 km/hr. The ratio of times taken by the passenger train to completely cross the freight train when: (i) they are moving in the same direction, and (ii) in the opposite direction is:

Show answer and solution

Answer: (c)

Distance to cross completely $= 60 + 120 = 180\text{ m}$ in both cases.
Same direction: $v_{rel} = 80 - 30 = 50\text{ km/h}$ ; opposite: $v_{rel} = 80 + 30 = 110\text{ km/h}$
$\frac{t_1}{t_2} = \frac{110}{50} = \frac{11}{5}$
So the answer is (c) $\frac{11}{5}$ .

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Q228NEET 2017Medium

Relative Velocity · Escalator problems

Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t1. On another day, if she remains stationary on the moving escalator, then the escalator takes her up in time t2. The time taken by her to walk up on the moving escalator will be

Show answer and solution

Answer: (b)

Length $L$ : her speed $\frac{L}{t_1}$ , escalator speed $\frac{L}{t_2}$
Walking on moving escalator: $t = \frac{L}{L/t_1 + L/t_2}$
$t = \frac{t_1t_2}{t_1 + t_2}$ , so the answer is (b).

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Q229JEE Main 2014 · 12 APR · SHIFT 1Medium

Relative Velocity · Escalator problems

A Person climbs up a stalled escalator in 60 s. If standing on the same but escalator running with constant velocity he takes 40 s. How much time is taken by the person to walk up the moving escalator?

Show answer and solution

Answer: (c)

Speeds add: $t = \frac{t_1t_2}{t_1 + t_2}$
$t = \frac{60 \times 40}{60 + 40} = 24\text{ s}$
So the answer is (c) 24 s.

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Q230NEET 2009 · AIPMTEasy

Relative Velocity · Relative velocity along a line

A bus is moving with a speed of 10 ms ${}^{-1}$ on a straight road. A scooterist wishes to overtake the bus in 100 s. If the bus is at a distance of 1 km from the scooterist, with what speed should the scooterist chase the bus?

Show answer and solution

Answer: (d)

Relative speed needed $= \frac{1000}{100} = 10\text{ m/s}$
$v - 10 = 10 \Rightarrow v = 20\text{ m/s}$
So the answer is (d) 20 m/s.

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Q231NEET 1988 · AIPMTEasy

Relative Velocity · Relative velocity along a line

A train of 150 metre length is going towards north direction at a speed of 10 m/s. A parrot flies at the speed of 5 m/s towards south direction parallel to the railways track. The time taken by the parrot to cross the train is

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Answer: (d)

Opposite directions: relative speed $= 10 + 5 = 15\text{ m/s}$
$t = \frac{150}{15} = 10\text{ s}$
So the answer is (d) 10 sec.

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