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Position-time and velocity-time graphs; speed, velocity and accelerated motion: NEET MCQs with answers

90 questions on this topic of Kinematics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A body starts from rest and moves with a uniform acceleration of 3 m/s². The distance it covers in the first 2 s is:

  1. 9 m
  2. 3 m
  3. 12 m
  4. 6 m
Show answer and explanation

Answer: (D) 6 m

Starting from rest, u = 0, so s = ut + ½at² = 0 + ½ × 3 × 2² = 6 m.

medium

2. A car moving at 108 km/h is brought to rest by a uniform retardation of 5 m/s². The stopping distance is:

  1. 6 m
  2. 1170 m
  3. 90 m
  4. 180 m
Show answer and explanation

Answer: (C) 90 m

Convert: v = 108/3.6 = 30 m/s. Using v² = u² − 2as with final speed 0: s = u²/2a = 30²/10 = 90 m.

hard

3. A metro train starts from station A from rest and accelerates at a uniform rate α = 2 m/s² until it reaches a speed v. It then immediately retards at a uniform rate β = 4 m/s² until it comes to rest at station B. If the total distance between the two stations is 600 m, the maximum speed v reached by the train is:

  1. 40 m/s
  2. 20 m/s
  3. 30 m/s
  4. 50 m/s
Show answer and explanation

Answer: (A) 40 m/s

For motion with uniform acceleration α followed immediately by deceleration β to rest, total distance S = v²/(2α) + v²/(2β) = (v²/2) × (1/α + 1/β) = (v²/2) × (1/2 + 1/4) = (v²/2) × (3/4) = 3v²/8. Given S = 600 m, we have 3v²/8 = 600 => v² = (600 × 8)/3 = 1600 => v = 40 m/s.

medium

4. A car moving at 90 km/h is brought to rest by a uniform retardation of 8 m/s². The stopping distance is:

  1. 39.1 m
  2. 506 m
  3. 3.13 m
  4. 78.1 m
Show answer and explanation

Answer: (A) 39.1 m

Convert: v = 90/3.6 = 25 m/s. Using v² = u² − 2as with final speed 0: s = u²/2a = 25²/16 = 39.1 m.

easy

5. A body starts from rest and moves with a uniform acceleration of 2 m/s². The distance it covers in the first 5 s is:

  1. 50 m
  2. 5 m
  3. 10 m
  4. 25 m
Show answer and explanation

Answer: (D) 25 m

Starting from rest, u = 0, so s = ut + ½at² = 0 + ½ × 2 × 5² = 25 m.

hard

6. A point moves along the x-axis with an initial velocity of 5 m/s. Its acceleration is given by a = -2v, where v is the instantaneous velocity. The distance traversed by the particle until its velocity reduces to 1 m/s is:

  1. 2.5 m
  2. 4 m
  3. 2 m
  4. 1.5 m
Show answer and explanation

Answer: (C) 2 m

Acceleration can be expressed as a = v(dv/dx). Given v(dv/dx) = -2v, so dv/dx = -2 (for v ≠ 0), which means dv = -2 dx. Integrating both sides: ∫ dv from 5 to 1 = -2 ∫ dx from 0 to s. Thus, 1 - 5 = -2s, giving -4 = -2s, or s = 2 m.

medium

7. A car moving at 54 km/h is brought to rest by a uniform retardation of 2 m/s². The stopping distance is:

  1. 7.5 m
  2. 729 m
  3. 56.3 m
  4. 113 m
Show answer and explanation

Answer: (C) 56.3 m

Convert: v = 54/3.6 = 15 m/s. Using v² = u² − 2as with final speed 0: s = u²/2a = 15²/4 = 56.3 m.

medium

8. A particle moves along the x-axis such that its position is given by x = 2t³ - 9t² + 12t - 5 (where x is in metres and t is in seconds). The acceleration of the particle when its velocity becomes zero for the second time is:

  1. +6 m/s²
  2. -6 m/s²
  3. +12 m/s²
  4. 0 m/s²
Show answer and explanation

Answer: (A) +6 m/s²

Velocity is v = dx/dt = 6t² - 18t + 12 = 6(t - 1)(t - 2). Thus, v = 0 at t = 1 s and t = 2 s. The second time is t = 2 s. Acceleration is a = dv/dt = 12t - 18. At t = 2 s, a = 12(2) - 18 = +6 m/s².

easy

9. A body starts from rest and moves with a uniform acceleration of 3 m/s². The distance it covers in the first 5 s is:

  1. 7.5 m
  2. 37.5 m
  3. 15 m
  4. 75 m
Show answer and explanation

Answer: (B) 37.5 m

Starting from rest, u = 0, so s = ut + ½at² = 0 + ½ × 3 × 5² = 37.5 m.

easy

10. A body starts from rest and moves with a uniform acceleration of 5 m/s². The distance it covers in the first 6 s is:

  1. 30 m
  2. 15 m
  3. 90 m
  4. 180 m
Show answer and explanation

Answer: (C) 90 m

Starting from rest, u = 0, so s = ut + ½at² = 0 + ½ × 5 × 6² = 90 m.

easy

11. Given below are two statements: Statement I: A particle can have a varying velocity while its speed remains constant. Statement II: A particle can have a varying speed while its velocity remains constant. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct.
  2. Statement I is correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are incorrect.
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect.

Velocity is a vector quantity (magnitude and direction), whereas speed is its magnitude. Direction can change while magnitude (speed) remains constant, as in uniform circular motion, so Statement I is correct. However, if speed changes, the magnitude of velocity changes, so velocity cannot remain constant; Statement II is incorrect.

easy

12. A body starts from rest and moves with a uniform acceleration of 4 m/s². The distance it covers in the first 8 s is:

  1. 16 m
  2. 128 m
  3. 256 m
  4. 32 m
Show answer and explanation

Answer: (B) 128 m

Starting from rest, u = 0, so s = ut + ½at² = 0 + ½ × 4 × 8² = 128 m.

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