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AS Daily A Level Physics question

2026-04-02 OCR A Mechanics & Materials (M3) 3.1.2 Motion — graphs of motion (displacement–time), interpreting gradient as velocity 3.1.2 Motion — average speed versus average velocity over a time interval

A walker moves along a straight footpath. Their displacement from a gate is recorded over 60 s and the displacement–time graph has three straight sections: from 0–20 s it increases uniformly from 0 m to +80 m; from 20–40 s it stays constant at +80 m; from 40–60 s it decreases uniformly to +20 m. Which statement must be true about the ratio (average speed) : (magnitude of average velocity) over the whole 60 s?

  1. A 1, because start and finish are both in front of the gate so distance and displacement are effectively the same.
  2. B 3.5, because stopping for 20 s halves the average speed compared with the outward motion.
  3. C 1/7, because the small net displacement makes the average velocity larger than the average speed.
  4. D 7, because the total distance is 140 m while the net displacement is 20 m over 60 s. (correct)

Answer

The correct answer is D.

Correct: D — 7, because the total distance is 140 m while the net displacement is 20 m over 60 s. A confuses distance with displacement: both positions are positive but the return leg means distance (80 + 0 + 60 = 140 m) is not equal to the 20 m displacement. B assumes the rest period simply halves the average speed and ignores that the ratio asked depends on distance versus displacement, not just the pause. C inverts the relationship and wrongly claims average velocity can exceed average speed; in fact average speed is always at least as large as the magnitude of average velocity. D uses the correct totals so the ratio is (140/60)/(20/60) = 140/20 = 7.

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