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AS Daily A Level Physics question

2026-02-23 OCR A Mechanics & Materials (M3) OCR-A 3.1.1 Kinematics — uniformly accelerated motion (suvat) OCR-A 3.1.1(b) Apply equations of motion to linear motion with constant acceleration

In a lab test, a dynamics trolley on a level track is slowed by a magnetic brake that provides nearly constant deceleration. From a launch speed of 1.5 m/s, it comes to rest in 1.2 m. If the same trolley–brake setup is used but the launch speed is increased to 3.0 m/s, what stopping distance should you predict?

  1. A 2.4 m
  2. B 3.6 m
  3. C 1.2 m
  4. D 4.8 m (correct)

Answer

The correct answer is D.

Correct: D — 4.8 m. With the same constant deceleration, stopping distance is proportional to the square of the launch speed, so doubling speed quadruples distance: 1.2 m × 4 = 4.8 m. A doubles the distance (2.4 m), a common linear-speed misconception ignoring that both stopping time and average speed double. B (3.6 m) assumes a non-quadruple scaling, mixing up proportionalities between speed, time, and distance. C (1.2 m) wrongly suggests no change, as if increased time were exactly offset by reduced average speed; under constant deceleration the average speed scales with initial speed, so distance cannot stay the same. D uses s ∝ u²: (3.0/1.5)² = 4.