Back to Daily Question archive

AS Daily A Level Physics question

2026-06-11 OCR A Kinematics (M3.1.1) OCR-A 3.1.1 Kinematics: velocity–time graphs (gradient = acceleration; area under graph = displacement)

In a lab, a motion sensor records a trolley that starts from rest, speeds up uniformly to 2.4 m/s in 2.0 s, then immediately slows uniformly to rest taking 4.0 s. The velocity–time graph is a straight-line rise followed by a straight-line fall back to zero. Which statement must be true?

  1. A The distance covered while slowing down is smaller than during speeding up.
  2. B The magnitudes of the accelerations in the two phases are equal because the speed change is the same.
  3. C The magnitude of the acceleration is greater while speeding up than while slowing down. (correct)
  4. D The total distance is equal to the maximum speed multiplied by the total time.

Answer

The correct answer is C.

Correct: C — The magnitude of the acceleration is greater while speeding up than while slowing down. A The deceleration phase lasts longer at moderate speeds, so the area under the graph (distance) is larger there, not smaller. B The speed changes by the same amount but over different times; different gradients on the velocity–time graph mean different acceleration magnitudes. C This follows because the same change in speed occurs in a shorter time during speeding up, so the gradient (and hence acceleration magnitude) is larger. D That would only be true for constant speed; here the triangular velocity–time shape means the distance is less than vmax × total time.

Try today's guest Daily MCQ A Level Physics MCQ practice Problem-solving previews