Motion In One Dimension JEE Main previous year questions with solutions

5 solved JEE Main questions on Motion In One Dimension, free to read — no sign-in needed. The full chapter has 178 questions; sign in to attempt the remaining 173 in the exam simulator.

1. Q1JEE Main 2026 (08 Apr, Shift 2)Motion Under Gravity
   A gas balloon is going up with a constant velocity of $10$ 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 $g=10$ m/s$^2$)
   1. A.$85$
   2. B.$150$
   3. C.$129$
   4. D.$125$
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   Answer: (D)

   When the stone is dropped, it acquires the initial velocity of the balloon. Taking the upward direction as positive, the initial velocity of the stone is $u = 10$ m/s. The displacement of the stone when it hits the ground is $S = -75$ m, and its acceleration is $a = -g = -10$ m/s$^2$. Using the equation of motion $S = ut + \dfrac{1}{2}at^2$: $-75 = 10t - \dfrac{1}{2}(10)t^2$ $-75 = 10t - 5t^2$ $5t^2 - 10t - 75 = 0$ $t^2 - 2t - 15 = 0$ $(t - 5)(t + 3) = 0$ Since time cannot be negative, $t = 5$ s. During this time, the balloon continues to move upwards with a constant velocity of $10$ m/s. The distance travelled by the balloon in $5$ s is: $d = v \times t = 10 \times 5 = 50$ m The total height of the balloon when the stone hits the ground is: $H = 75 + 50 = 125$ m Answer: $125$
2. Q2JEE Main 2025 (08 Apr, Shift 2)Non-uniform Motion
   A body of mass 2 kg moving with velocity of $\overrightarrow{\mathrm{v}}_{\mathrm{in}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}} \mathrm{ms}^{-1}$ enters into a constant force field of 6 N directed along positive z -axis. If the body remains in the field for a period of $\frac{5}{3}$ seconds, then velocity of the body when it emerges from force field is
   1. A.$4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$
   2. B.$3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$
   3. C.$3 \hat{i}+4 \hat{j}-5 \hat{k}$
   4. D.$3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\sqrt{5} \hat{\mathrm{k}}$
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   Answer: (B)

   $\begin{aligned} & \overrightarrow{\mathrm{a}}=\frac{\mathrm{B}}{2} \hat{\mathrm{k}}=3 \hat{\mathrm{k}}, \mathrm{t}=\frac{5}{3} \mathrm{~s} \\ & \overrightarrow{\mathrm{u}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{u}}+\overrightarrow{\mathrm{a}} \mathrm{t}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\end{aligned}$
3. Q3JEE Main 2024 (08 Apr, Shift 1)Rest and Motion
   A clock has $75 \mathrm{~cm}, 60 \mathrm{~cm}$ long second hand and minute hand respectively. In 30 minutes duration the tip of second hand will travel $x$ distance more than the tip of minute hand. The value of $x$ in meter is nearly (Take $\pi=3.14$ ) :
   1. A.140.5
   2. B.118.9
   3. C.139.4
   4. D.220.0
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   Answer: (C)

   $\begin{aligned} \mathrm{x}_{\min }= & \pi \times \mathrm{r}_{\min } \\ = & \pi \times \frac{60}{100} \mathrm{~m} . \\ \mathrm{x}_{\text {second }} & =30 \times 2 \pi \times \mathrm{r}_{\text {second }} \\ & =30 \times 2 \pi \times \frac{75}{100} \\ \mathrm{x}= & \mathrm{x}_{\text {second }}-\mathrm{x}_{\min } \\ = & 139.4 \mathrm{~m}\end{aligned}$
4. Q4JEE Main 2023 (08 Apr, Shift 2)Graphs of motion in one dimension
   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.
   1. A.Both Statement I and Statement II are true
   2. B.Both Statement I and Statement II are false
   3. C.Statement I is correct but Statement II is false
   4. D.Statement I is incorrect but Statement II is true
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   Answer: (D)

   The definition of displacement is given by the product of velocity and time. $\int vdt=∆x=$displacement of body in given time. Distance is a scalar quantity and displacement is a vector quantity defined by the shortest distance between the initial and the final position. Acceleration is defined by the rate of change of velocity. $\int adt=∆v=$change in velocity in given time. Thus, the area under the velocity time graph gives the displacement of the body and the area under the acceleration time graph gives the change in velocity of the body.
5. Q5JEE Main 2022 (27 Jul, Shift 2)Uniform Motion
   The velocity of the bullet becomes one third after it penetrates $4cm$ 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 $\left(4+x\right)cm$ inside the block. The value of $x$ is
   1. A.$2.0$
   2. B.$1.0$
   3. C.$0.5$
   4. D.$1.5$
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   Answer: (C)

   According to the given information for the first $4cm$, $u={v}_{0}$ and $v=\dfrac{{v}_{0}}{3}$. Applying equation of motion for constant acceleration, ${v}^{2}={u}^{2}+2as$ $\Rightarrow {\left(\dfrac{{v}_{0}}{3}\right)}^{2}={v}^{2}+2a\left(4\right) \Rightarrow a=-\dfrac{8{{v}_{0}}^{2}}{9\left(8\right)}=-\dfrac{{{v}_{0}}^{2}}{9}$ When the body stops, $v=0$ $\Rightarrow 0={{v}_{0}}^{2}+2a\left(4+x\right)$ $\Rightarrow {{v}_{0}}^{2}=2\left(\dfrac{{{v}_{0}}^{2}}{9}\right)\left(4+x\right)$ $\Rightarrow 4.5=4+x$ $\Rightarrow x=0.5cm$