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A2 Daily A Level Physics question

2026-05-01 OCR A Foundations of Physics (M2) OCR A H556 Module 3: Forces and motion — 3.1 Motion (kinematics): uniform acceleration (equations of motion) OCR A H556 Module 3: Forces and motion — velocity–time graph reasoning (average speed during constant deceleration)

On a straight test track, the same car is braked with the same constant braking force on level dry tarmac. In run 1 the driver begins braking at 20 m s−1; in run 2 the driver begins braking at 40 m s−1. Air resistance is small compared with the braking force so the deceleration is essentially the same in both runs. Compared with run 1, how do the stopping distance and stopping time change in run 2?

  1. A Stopping distance doubles; stopping time doubles.
  2. B Stopping distance doubles; stopping time is unchanged.
  3. C Stopping distance becomes four times larger; stopping time is unchanged.
  4. D Stopping distance becomes four times larger; stopping time doubles. (correct)

Answer

The correct answer is D.

Correct: D — Stopping distance becomes four times larger; stopping time doubles. A assumes both scale directly with speed; however with the same deceleration the time doubles while the average speed during braking also doubles, so distance scales with the square, not linearly. B is wrong because with the same deceleration, cutting the speed to zero takes twice as long when the initial speed is doubled. C is wrong because although the distance does become four times larger, the time cannot stay the same when the deceleration is fixed; it must double. D matches constant-deceleration reasoning: speed falls linearly to zero so average speed doubles and the time to stop doubles, giving distance = average speed × time that is four times larger.