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AS Daily A Level Physics question

2025-09-12 OCR A Physical quantities & SI (M2), Scalar vs vector (M2), Kinematics: displacement–time (M3) 3.1.1(c)

A trolley on a straight track is monitored. Its displacement from the start increases uniformly from 0.0 m to +8.0 m during the first 10 s. It then decreases uniformly, reaching −4.0 m at t = 25 s. Which statement must be true?

  1. A The trolley’s constant speed is the same in both intervals (0.80 m s^-1), but in opposite directions. (correct)
  2. B The trolley’s speed is greater in the second interval because the displacement change is larger (12 m vs 8 m).
  3. C The average velocity over the whole 25 s is zero, since it reversed direction and crossed x = 0 m.
  4. D The average speed over the whole 25 s is 0.64 m s^-1 by averaging the two interval speeds.

Answer

The correct answer is A.

Correct: A — The trolley’s constant speed is the same in both intervals (0.80 m s^-1), but in opposite directions. 8.0 m in 10 s gives 0.80 m s^-1 outward; 12.0 m in 15 s gives 0.80 m s^-1 back, so equal magnitudes with opposite directions. B is wrong because it compares distances but ignores the longer time (15 s) in the second interval; speed depends on both distance and time. C is wrong because average velocity is total displacement over total time: (−4.0 m)/25 s = −0.16 m s^-1, not zero. D is wrong because simply averaging interval speeds (and here miscalculating) is invalid for unequal time intervals; the correct average speed is total distance/total time = (8.0 + 12.0) m / 25 s = 0.80 m s^-1.

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