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AS Daily A Level Physics question

2025-11-26 OCR A High level GCSE physics — kinematics (velocity–time graphs) 3.1.1(d)

A lab trolley moves in a straight line. Its velocity changes uniformly from +4.0 m/s at t = 0.0 s to −2.0 m/s at t = 3.0 s. Take forward as positive. Which statement must be true about its position relative to the start at t = 3.0 s?

  1. A Equal forward and backward distances are covered in the 3.0 s, so it is back at the start.
  2. B It is 1.0 m behind the start because the final velocity is negative, so the 3.0 s average velocity must be negative.
  3. C It is 3.0 m ahead of the start because the positive area 0–2 s is 0.5×2×4 = 4 m and the negative area 2–3 s is 0.5×1×2 = 1 m. (correct)
  4. D Continuing the same acceleration would bring it back to the start at t = 4.0 s; therefore by t = 3.0 s it must already be behind the start.

Answer

The correct answer is C.

Correct: C — It is 3.0 m ahead of the start because the positive area 0–2 s is 0.5×2×4 = 4 m and the negative area 2–3 s is 0.5×1×2 = 1 m.. A assumes equal distances from unequal speeds and times; in fact the positive area (4 m) exceeds the negative area (1 m). B treats the sign of the final velocity as fixing the average; with a uniform change the mean velocity is positive, so the displacement is forward, not −1 m. C correctly uses areas under the velocity–time graph with sign to find a net +3 m displacement. D notes a valid limiting case (return at 4.0 s when areas cancel) but wrongly infers that at 3.0 s the trolley is already behind; by 3.0 s only 1 m of backward area has accumulated, so it remains ahead.

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