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AS Daily A Level Physics question

2026-05-14 OCR A Foundations of Physics (M2) OCR-A Module 3.1.1 Motion in a straight line: displacement–time graphs; gradient as velocity (sign and magnitude) OCR-A Module 2.3.1 Scalars and vectors: distinction between speed (scalar) and velocity (vector)

In a lab tracking experiment, the displacement s of a trolley along a straight track is plotted against time t. The graph has three sections: 0–3 s: s increases linearly from 0 m to +6 m; 3–5 s: s remains constant at +6 m; 5–9 s: s decreases smoothly from +6 m to –2 m and the curve gets progressively steeper as time increases. Which statement must be true?

  1. A The trolley’s average velocity over 0–9 s is zero because it finishes near where it started.
  2. B Between 5 s and 9 s the trolley is moving toward its starting point and speeding up. (correct)
  3. C The total distance travelled from 0 to 9 s is 8 m.
  4. D During 3–5 s the trolley’s speed is negative because the graph is flat.

Answer

The correct answer is B.

Correct: B — Between 5 s and 9 s the trolley is moving toward its starting point and speeding up. A The average velocity depends on net displacement, which is −2 m over 9 s, so it is not zero; finishing near the start does not make the average velocity zero. B This interval shows displacement decreasing with an increasingly steep slope, so velocity is negative (toward the start) and the speed increases. C This confuses distance with displacement: the trolley goes +6 m out and then 8 m back, so the total distance is greater than 8 m. D Speed is a scalar and cannot be negative; a flat displacement–time graph means zero velocity and therefore zero speed.

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