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AS Daily A Level Physics question

2026-03-02 OCR A Kinematics (velocity–time graphs, acceleration) Module 3: Forces and motion — 3.1.2 Motion (velocity–time graphs; acceleration; displacement as area under v–t graph) Module 3: Forces and motion — 3.1.2 Motion (uniform acceleration; equations of motion, qualitative use)

On a test track, two cars brake uniformly over the same 6 s interval. Car A's speed falls linearly from 12 m/s to 0 m/s. Car B's speed falls linearly from 16 m/s to 4 m/s. Over this 6 s, what is the ratio of the distances they travel (Car B : Car A)?

  1. A 4/3
  2. B 5/3 (correct)
  3. C 1
  4. D 2

Answer

The correct answer is B.

Correct: B — 5/3. For a straight-line decrease in speed, distance in a given time is the area under the velocity–time graph (mean of start and end speeds × time): A = ((12 + 0)/2) × 6 = 36 m, B = ((16 + 4)/2) × 6 = 60 m, so B:A = 60:36 = 5:3. A 4/3 uses only the initial speed ratio (16/12) and ignores that B still has 4 m/s at the end, which increases the area (distance). B is correct because the trapezium areas (or mean speeds 10 m/s vs 6 m/s) give a ratio of 10/6 = 5/3. C 1 assumes equal deceleration and equal time imply equal distance, but distance depends on average speed, which differs here. D 2 likely confuses peak speed with average speed or double-counts part of the area; the correct average-speed ratio is 10/6, not 12/6 or 16/8.

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