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A2 Daily A Level Physics question

2026-05-13 OCR A Kinematics (displacement–time) OCR A Physics A Module 3.1.1 Kinematics — displacement–time graphs; speed vs velocity; distance vs displacement

In a lab, a trolley moves along a straight track while a motion sensor records its displacement from the start against time. The displacement–time graph is piecewise straight: from 0–6 s it rises linearly; from 6–8 s it is horizontal; from 8–12 s it falls linearly with a steeper gradient in magnitude than before, finishing at a negative displacement relative to the starting position. Which statement must be true?

  1. A Over the whole 12 s, the trolley’s average speed equals the magnitude of its average velocity.
  2. B The trolley’s greatest speed occurs during the horizontal section of the graph (6–8 s).
  3. C The trolley’s acceleration is constant and negative throughout the motion.
  4. D The total distance travelled is greater than the magnitude of the final displacement from the start. (correct)

Answer

The correct answer is D.

Correct: D — The total distance travelled is greater than the magnitude of the final displacement from the start. A change in gradient from positive to negative shows a reversal in direction, so the path length (out and back) must exceed the net displacement. A is wrong because reversing direction means average speed exceeds the magnitude of average velocity. B is wrong because a horizontal displacement–time segment has zero gradient, so the trolley is stationary (speed zero), not at its greatest speed. C is wrong because the straight-line segments indicate constant velocity (zero acceleration), and the acceleration changes at the joins; it cannot be constant and negative throughout. D is correct for any motion that includes a reversal: distance accumulates while the net displacement can partly cancel.