Motion in a Straight Line NEET PYQs (2005–2026) with Solutions

10 solved Motion in a Straight Line PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumAverage speed and average velocity

   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 $\alpha = 1\ \mathrm{ms^{-2}}$, $\beta = 6\ \mathrm{ms^{-1}}$ and $\gamma = 5\ \mathrm{m}$. The average speed of the particle, in $\mathrm{ms^{-1}}$ from $t = 0$ to $t = 6\ \mathrm{s}$ is:

   1. AOption A: $12$
   2. BOption B: $6$
   3. COption C: $3$
   4. DOption D: $0$

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   Correct answer: (C) $3$

   Explanation

   $s(t) = t^2 - 6t + 5$, so the velocity is $v = \frac{ds}{dt} = 2t - 6$. It is zero at $t = 3\ \mathrm{s}$, where the particle reverses direction. $s(0) = 5\ \mathrm{m}$, $s(3) = 9 - 18 + 5 = -4\ \mathrm{m}$, $s(6) = 36 - 36 + 5 = 5\ \mathrm{m}$. Distance travelled $= |{-4} - 5| + |5 - (-4)| = 9 + 9 = 18\ \mathrm{m}$. Average speed $= \frac{18\ \mathrm{m}}{6\ \mathrm{s}} = 3\ \mathrm{ms^{-1}}$, option (C). (The average velocity is 0, which is the trap in option (D).)

2. Question 2 (NEET 2026)

   NEET 2026EasyMotion Graphs

   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?

   1. AOption A: C only
   2. BOption B: D only
   3. COption C: B only
   4. DOption D: A and E only

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   Correct answer: (A) C only

   Explanation

   For a ball thrown vertically upward and returning under gravity, acceleration remains constant and downward throughout the motion: Acceleration = −g Therefore the slope of the velocity-time graph must remain constant and negative during the entire journey. Hence velocity decreases linearly with time: - Positive during upward motion - Zero at highest point - Negative during downward motion Thus the correct graph is a straight line with constant negative slope crossing the time axis. Only graph C satisfies this condition. Hence option (A) is correct.

3. Question 3 (NEET 2026)

   NEET 2026EasyKinematic Equations

   When a ruler falls vertically, 5 different persons catch it with different reaction times (g = 9.8 m s⁻²): A. Person A has reaction time 0.20 s B. Person B has reaction time 0.22 s C. Person C has reaction time 0.18 s D. Person D has reaction time 0.19 s E. Person E has reaction time 0.21 s What is the correct order of the distance travelled by the ruler for each person?

   1. AOption A: B > E > A > C > D
   2. BOption B: C > D > A > B > E
   3. COption C: B > E > A > D > C
   4. DOption D: C > D > A > E > B

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   Correct answer: (C) B > E > A > D > C

   Explanation

   Distance covered in free fall is proportional to t². Hence larger reaction time gives larger distance. Ordering reaction times: 0.22 > 0.21 > 0.20 > 0.19 > 0.18 ⇒ B > E > A > D > C. Therefore option (C) is correct.

4. Question 4 (NEET 2025)

   NEET 2025MediumKinematic Equations

   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

   1. AOption A: $-\dfrac{2}{(x+2)^3}$
   2. BOption B: $-\dfrac{2}{(2x+1)^3}$
   3. COption C: $+\dfrac{2}{(x+1)^3}$
   4. DOption D: $+\dfrac{2}{2x+1}$

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   Correct answer: (B) $-\dfrac{2}{(2x+1)^3}$

   Explanation

   Given, $$t = x^2 + x$$ Differentiate with respect to $x$:$$\frac{dt}{dx} = 2x + 1$$Hence,$$v = \frac{dx}{dt} = \frac{1}{2x+1}$$Acceleration is given by$$a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt}$$Now,$$\frac{dv}{dx} = -\frac{2}{(2x+1)^2}$$Therefore,$$a = -\frac{2}{(2x+1)^2}\times\frac{1}{2x+1}$$$$a = -\frac{2}{(2x+1)^3}$$ Hence, the correct option is (2).

5. Question 5 (NEET 2024)

   NEET 2024MediumMotion 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:

   1. AOption A: Graph (1)
   2. BOption B: Graph (2)
   3. COption C: Graph (3)
   4. DOption D: Graph (4)

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   Correct answer: (A) Graph (1)

   Explanation

   Acceleration is the slope of the velocity-time graph. Initially, slope is positive constant, then zero during constant velocity, and finally negative constant while velocity decreases. Thus the acceleration-time graph consists of positive acceleration, then zero acceleration, and then negative acceleration. Hence, option (A) is correct.

6. Question 6 (NEET 2023)

   NEET 2023MediumSpeed & Velocity

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

   1. AOption A: Option 1
   2. BOption B: Option 2
   3. COption C: Option 3
   4. DOption D: Option 4

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   Correct answer: (C) Option 3

   Explanation

   Let total distance = 2d. Time for first half: t₁ = d/v Time for second half: t₂ = d/2v Total time: t = d/v + d/2v = 3d/2v Average speed: v_avg = total distance / total time = 2d / (3d/2v) = 4v/3 Hence, the average speed is 4v/3.

7. Question 7 (NEET 2023)

   NEET 2023EasyMotion in Straight Line

   A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity 4 m s⁻¹. The ball strikes the water surface after 4 s. The height of bridge above water surface is (Take g = 10 m s⁻²):

   1. AOption A: 56 m
   2. BOption B: 60 m
   3. COption C: 64 m
   4. DOption D: 68 m

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   Correct answer: (C) 64 m

   Explanation

   Using the equation of motion: s = ut + (1/2)at^2 Taking upward direction as positive: u = 4 m/s a = -10 m/s^2 t = 4 s s = 4(4) + (1/2)(-10)(4)^2 s = 16 - 80 s = -64 m Hence, the bridge is 64 m above the water surface.

8. Question 8 (NEET 2022)

   NEET 2022EasyKinematic Equations

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

   1. AOption A: 1 : 1 : 1 : 1
   2. BOption B: 1 : 2 : 3 : 4
   3. COption C: 1 : 4 : 9 : 16
   4. DOption D: 1 : 3 : 5 : 7

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   Correct answer: (D) 1 : 3 : 5 : 7

   Explanation

   For a freely falling body under constant acceleration, distance covered in the nth second is: sₙ = u + a/2 (2n − 1) For free fall, u = 0 and a = g: sₙ ∝ (2n − 1) Thus, distances in successive seconds are: 1st : 2nd : 3rd : 4th = 1 : 3 : 5 : 7 Hence, option (D) is correct.

9. Question 9 (NEET 2022)

   NEET 2022EasyMotion Graphs

   The displacement-time graphs of two moving particles make angles of 30° and 45° with the x-axis as shown in the figure. The ratio of their respective velocities is:

   1. AOption A: 1 : √3
   2. BOption B: √3 : 1
   3. COption C: 1 : 1
   4. DOption D: 1 : 2

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   Correct answer: (A) 1 : √3

   Explanation

   Velocity equals slope of displacement-time graph. v₁/v₂ = tan30°/tan45° = (1/√3)/1 = 1 : √3 Hence option (A) is correct.

10. Question 10 (NEET 2021)

    NEET 2021MediumKinematic Equations

    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 Sₙ / Sₙ₊₁ is:

    1. AOption A: 2n / (2n − 1)
    2. BOption B: (2n − 1) / 2n
    3. COption C: (2n − 1) / (2n + 1)
    4. DOption D: (2n + 1) / (2n − 1)

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    Correct answer: (C) (2n − 1) / (2n + 1)

    Explanation

    For motion down a smooth incline, acceleration a = gsinθ. Distance travelled in nth second: Sₙ = u + (a/2)(2n−1) Since u = 0, Sₙ = (gsinθ/2)(2n−1) Similarly, Sₙ₊₁ = (gsinθ/2)(2n+1) Therefore, Sₙ/Sₙ₊₁ = (2n−1)/(2n+1).