Motion in a Straight Line

9 NEET-UG practice questions on Motion in a Straight Line, part of Physics. The first 9 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Medium

   The position of a particle is x = 2t³ − 9t² + 12t + 1, with x in metres and t in seconds. What is its acceleration at the instant its velocity becomes zero for the second time?

   1. A+6 m/s²
   2. BZero
   3. C−6 m/s²
   4. D+24 m/s²

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   Answer: A · +6 m/s²

   v = dx/dt = 6t² − 18t + 12 = 6(t − 1)(t − 2), so v = 0 at t = 1 s and t = 2 s. a = dv/dt = 12t − 18, and at t = 2 s this is +6 m/s².

   • B: This assumes acceleration vanishes when velocity does.
   • C: This is the acceleration at the first stop, t = 1 s.
   • D: This leaves out the −18 term of the acceleration.

   Remember this

   Zero velocity does not mean zero acceleration; differentiate first, then substitute the time.

2. Question 2 · Easy

   Match List-I with List-II. List-I (feature of a graph) A. Slope of the x-t graph B. Slope of the v-t graph C. Area under the v-t graph D. Area under the a-t graph List-II (what it gives) I. Displacement II. Change in velocity III. Velocity IV. Acceleration

   1. AA-III, B-IV, C-I, D-II
   2. BA-III, B-IV, C-II, D-I
   3. CA-IV, B-III, C-I, D-II
   4. DA-I, B-IV, C-III, D-II

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   Answer: A · A-III, B-IV, C-I, D-II

   Velocity is dx/dt and acceleration is dv/dt, so these are slopes. Displacement is the integral of v dt and change in velocity is the integral of a dt, so these are areas.

   • B: This swaps the two areas.
   • C: This swaps the two slopes.
   • D: This swaps the slope of the x-t graph with the area under the v-t graph.

   Remember this

   Slope differentiates and area integrates, moving one step along x, v, a.

3. Question 3 · Hard

   A particle starts from rest and moves along a line with acceleration a = (6 − 2t) m/s², t in seconds. How far has it travelled when it next comes to rest?

   1. A27 m
   2. B36 m
   3. C108 m
   4. D72 m

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   Answer: B · 36 m

   v = 6t − t², which is zero again at t = 6 s, and v stays positive in between. x = 3t² − t³/3 = 108 − 72 = 36 m. Check: the v-t curve is a parabola with peak 9 m/s, and its area is (2/3)(9)(6) = 36 m.

   • A: This treats the curved v-t graph as a triangle.
   • C: This keeps only the 3t² term of the displacement.
   • D: This keeps only the t³/3 term of the displacement.

   Remember this

   With time-dependent acceleration, integrate twice; the constant-acceleration equations do not apply.

4. Question 4 · Medium

   Starting from rest, a lift's speed rises uniformly to 4 m/s in 2 s. The lift then runs at this speed for 5 s and slows uniformly to rest in another 3 s, all in one direction. What is its average velocity for the whole trip?

   1. A3 m/s
   2. B4 m/s
   3. C2 m/s
   4. D2.67 m/s

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   Answer: A · 3 m/s

   Displacement is the area of the v-t trapezium: 4 + 20 + 6 = 30 m. Check: (1/2)(5 + 10)(4) = 30 m. Average velocity = 30/10 = 3 m/s.

   • B: This is the peak speed, not the average.
   • C: This is the average during the speeding-up stage alone.
   • D: This takes the mean of the three stage averages without weighting by time.

   Remember this

   Average velocity = total displacement/total time, never the mean of the speeds in the separate stages.

5. Question 5 · Medium

   A ball is thrown straight up from the ground at 30 m/s. At the same instant, a second ball is released from rest at a point 60 m directly above the first. Taking g = 10 m/s², at what height above the ground do they meet?

   1. A40 m
   2. B60 m
   3. C20 m
   4. D30 m

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   Answer: A · 40 m

   Both balls have the same acceleration, so their relative speed stays 30 m/s and they meet at t = 60/30 = 2 s. Height of the first ball = 30 × 2 − 5 × 4 = 40 m. Check: the dropped ball falls 5 × 4 = 20 m, and 60 − 20 = 40 m.

   • B: This is the starting height of the dropped ball.
   • C: This is how far the dropped ball falls, not the meeting height.
   • D: This assumes they meet halfway.

   Remember this

   Two bodies in free fall have zero relative acceleration, so their relative velocity stays constant.

6. Question 6 · Hard

   Starting from rest at the origin, a particle moves along a line with the acceleration-time graph shown. How far is it from the origin at the instant its speed is greatest?

   

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   1. A13.5 m
   2. B18 m
   3. C27 m
   4. D9 m

   Answer and explanation in LearnFree account

7. Question 7 · Medium

   Position of a runner on a straight road is plotted against time, with +x as the positive direction. Over the full 10 s, what are the average speed and the average velocity?

   

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   1. A6.25 m/s and −1 m/s
   2. B1 m/s and −1 m/s
   3. C5 m/s and −1 m/s
   4. D5 m/s and +1 m/s

   Answer and explanation in LearnFree account

8. Question 8 · Easy

   A cart moves along a straight track without reversing. Using its velocity-time graph, find the distance it covers in the 8 s shown.

   

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   1. A54 m
   2. B66 m
   3. C96 m
   4. D48 m

   Answer and explanation in LearnFree account

9. Question 9 · Medium

   From rest, a scooter on a straight road has acceleration +2 m/s² for the first 2 s, zero for the next 2 s and −2 m/s² for the last 2 s. Its velocity against time is correctly shown by which graph?

   

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   1. AGraph B
   2. BGraph D
   3. CGraph A
   4. DGraph C

   Answer and explanation in LearnFree account