A2 Daily A Level Physics question
On a dry test track, the same car makes two emergency stops under identical conditions so its deceleration is effectively constant and unchanged on both runs. On the second run the initial speed is 30% higher than on the first. Compared with the first stop, what happens to the stopping distance and the stopping time?
Answer
The correct answer is B.
Correct: B — Stopping distance increases by about 70%; stopping time increases by about 30%. With the same constant deceleration, stopping time is initial speed divided by deceleration, so it rises in direct proportion (30% higher). The stopping distance is average speed × time; with final speed zero, average speed is half the initial speed, so distance scales with the square of initial speed: 1.3^2 ≈ 1.69 (about 70% higher). A assumes distance scales linearly with speed, but distance scales with speed squared under constant deceleration. B matches s ∝ u^2 (≈1.69×) and t ∝ u (≈1.3×). C incorrectly makes stopping time rise like speed squared; time is only proportional to speed when deceleration is unchanged. D swaps the scalings, giving too small a change in distance and too large a change in time.