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AS Daily A Level Physics question

2026-03-20 OCR A Kinematics (M3.1.1) OCR-A 3.1.1 Kinematics: velocity–time graphs; area under v–t gives displacement OCR-A 3.1.2 Motion with uniform acceleration (suvat) — uniform acceleration represented by straight-line v–t graphs

In a lab test, two trolleys move along a straight track for 10 s. Trolley A moves at a constant 6 m/s. Trolley B starts from rest and increases speed uniformly, reaching 12 m/s at 10 s. Which statement about the distances travelled in the first 10 s must be true?

  1. A Trolley A travels 20 m further than trolley B in 10 s, because A’s speed is higher early on.
  2. B Trolley B travels 20 m further than trolley A in 10 s, because B reaches a higher final speed.
  3. C They travel the same distance in 10 s (each 60 m), because the areas under their velocity–time graphs are equal. (correct)
  4. D Trolley B travels half as far as trolley A in 10 s, because its speed starts from zero.

Answer

The correct answer is C.

Correct: C — They travel the same distance in 10 s (each 60 m), because the areas under their velocity–time graphs are equal. A: Wrong, because A’s distance is 6×10 = 60 m while B’s is the triangular area ½×10×12 = 60 m, not 40 m. B: Wrong, a higher final speed does not guarantee a greater distance; B’s average speed is ½×12 = 6 m/s, giving the same 60 m as A. C: Correct, equal areas (rectangle vs triangle) both give 60 m over 10 s. D: Wrong, half as far would be 30 m, but B’s average speed is 6 m/s so it also covers 60 m.

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