Motion in a Straight Line JEE Main PYQs (2020–2026)

Solved Motion in a Straight Line PYQs

3 recent questions with answers. The full set of 91 is free to practise.

1. JEE Main 2026EasyMotion under Gravity
   Q1

   A gas balloon is going up with a constant velocity of $10m/s$ . 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 $g=10m/s^2$ )

   • A85
   • B150
   • C129
   • D125

   Show answer & solution

   Correct answer: (D)

   When an object is dropped from a moving carrier (like the balloon), it gets the velocity of that carrier at the exact moment of release.
   
   Initial velocity of the stone (balloon) is $u_s=10m/s$ (upwards).
   
   Initial height of the stone is $h=75m$ .
   
   The acceleration due to gravity is $g=-10m/s^2$ (downwards).
   
   Using the second equation of motion for the stone’s displacement along vertical direction,
   
   $$y=u_{y}t+\frac{1}{2}{a}_y{t}^2$$
   
   Since the stone eventually hits the ground (at height 0 ), its displacement from the starting point ( 75 m ) is $y=$ -75 m .
   
   $$-75=(10)t+\frac{1}{2}(-10)t^2$$
   
   $$\Rightarrow -75=10t-5t^2$$
   
   $$\Rightarrow 15=-2t+t^2$$
   
   $$\Rightarrow t^2-2t-15=0$$
   
   $$\Rightarrow (t-5)(t+3)=0$$
   
   Since time cannot be negative, we take $t=5$ seconds.
   
   While the stone is falling for 5 seconds, the balloon continues to move upward at a constant velocity.
   
   The constant velocity of balloon is $v_b=10m/s$ .
   
   The time interval of journey is $t=5$ seconds.
   
   Displacement $=$ Velocity × Time
   
   Displacement $=10\times 5=50m$
   
   The final height is the initial height plus the displacement covered during the 5 seconds.
   
   Total Height $=75m+50m=125m$
   
   Therefore, the height of the balloon when the stone hits the ground is 125 m .
   
   Hence, the correct option is (D).
2. JEE Main 2026HardInstantaneous Velocity, Acceleration and Calculus-Based Kinematics
   Q2

   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 :

   • AA, C, D Only
   • BA, B, C Only
   • CC, E Only
   • DA, C Only

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   Correct answer: (B)

   The original position at $t=0$ is $x(0)=4(0{)}^3-3(0)=0$ .
   
   To find when it returns to the origin :
   
   $$4t^3-3t=0$$
   
   $$\Rightarrow t\left(4t^2-3\right)=0$$
   
   $$4t^2=3\Rightarrow t^2=0.75\Rightarrow t=\sqrt{0.75}\approx 0.866s$$
   
   Hence, the Statement A is correct.
   
   A turning point occurs when velocity $v(t)=0$ .
   
   $$v(t)=\frac{dx}{dt}=12t^2-3$$
   
   At turning point,
   
   $$v(t)=0\Rightarrow 12t^2=3\Rightarrow t^2=0.25\Rightarrow t=0.5s$$
   
   So, the position at turning point is,
   
   $$x(0.5)=4(0.5{)}^3-3(0.5)=4(0.125)-1.5=0.5-1.5=-1.0m$$
   
   The distance from the origin is $|-1.0|=1$ unit.
   
   Hence, the statements B is correct and statement D is incorrect. Statement E is incorrect because the particle does turn back at $t=0.5s$ .
   
   Acceleration $a(t)$ is the derivative of velocity:
   
   $$a(t)=\frac{dv}{dt}=24t$$
   
   Since the particle starts moving from $t=0$ , for all $t\ge 0$ , the value of 24 t will always be greater than or equal to zero.
   
   Hence, statement C is correct.
   
   Therefore, statements $A,B$ , and C and D are correct. The option 2 best represents the correct set.
   
   Hence, the correct option is (B).
3. JEE Main 2025HardRelative Velocity in One Dimension
   Q3

   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

   Show answer & solution

   Correct answer: 3

   Here, we will use the concept of relative motion.
   
   Let initial (at $t=0$ ) position of both cars is x = 0 (As they cross each other at t = 0 for Ist time)
   
   given : For P
   
   $$a_P\propto t$$
   
   For Q,
   
   $a_Q=a^{\prime }$ (Let) (constant)
   
   $\Rightarrow a_{p^{(t)}}=kt$ where, k = constant
   
   $\Rightarrow \frac{d{v}_p(t)}{dt}=kt$ (as $a=\frac{dv}{dt}$ )
   
   $\Rightarrow {\int }_0^{v_p(t)}d{v}_p(t)={\int }_0^{t}ktdt$ [let p and Q both starts from rest]
   
   $$\Rightarrow v_p(t)=\frac{k{t}^2}{2}$$
   
   $\Rightarrow \frac{d{x}_p(t)}{dt}=\frac{k{t}^2}{2}$ [As $v=\frac{dx}{dt}$ ]
   
   $$\Rightarrow {\int }_0^{x_p(t)}d{x}_{p(t)}={\int }_0^t\frac{k{t}^2}{2}dt$$
   
   $\Rightarrow x_p(t)=\frac{k{t}^3}{6}$ …. (1)
   
   Now, for Q,
   
   $x_Q(t)=\frac{1}{2}{a}^1{t}^2$ …. (2) [using Newton’s 2nd equation of motion]
   
   So, relative position of P w.r.t. Q,
   
   $$x_{P{Q}^{(t)}}=x_p(t)-x_Q(t)$$
   
   $\Rightarrow x_{P{Q}^{(t)}}=\frac{k{t}^3}{6}-\frac{1}{2}{a}^1{t}^2$ (From (1) and (2))
   
   When both cars cross each other, $X_{PQ}=O$
   
   As $X_{PQ}(t)$ is a cubic polynomial, so it has maximum 3 roots.
   
   Hence, the maximum number of crossing = 3

Topic-wise question count

Where JEE Main has asked from Motion in a Straight Line, most frequent first.

• Motion under Gravity23
• Instantaneous Velocity, Acceleration and Calculus-Based Kinematics19
• Graphical Representation of Motion18
• Kinematic Equations for Uniformly Accelerated Motion14
• Distance, Displacement, Average Speed and Average Velocity11
• Relative Velocity in One Dimension6

Year-wise Motion in a Straight Line questions in JEE Main

Number of questions from this chapter in each year's paper(s).