A passenger train travels between two stations. It leaves the first station from rest and accelerates uniformly. It then travels at a constant speed before decelerating uniformly to rest at the second station. Describe the motion of the train during each stage of the journey, explaining how the distance travelled during the acceleration stage compares to the distance travelled during the deceleration stage if both stages last the same amount of time but the deceleration is twice the magnitude of the acceleration.

Edexcel GCSE Physics (1PH0) — 2.1 Motion · Describe · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

A passenger train travels between two stations. It leaves the first station from rest and accelerates uniformly, then travels at a constant speed, before decelerating uniformly to rest at the second station. The deceleration stage lasts the same length of time as the acceleration stage, but the magnitude of the deceleration is twice the magnitude of the acceleration.

Model answer (4 marks)

During the acceleration stage the train starts from rest and its speed increases at a constant rate – the velocity rises uniformly from 0 to the top speed.
During the constant‑speed stage the train moves with the same speed; its velocity is unchanged and the acceleration is zero.
During the deceleration stage the train’s speed decreases at a constant rate – the velocity falls uniformly from the top speed to 0.
Because the deceleration is twice the magnitude of the acceleration but the two stages last the same time, the top speed reached during acceleration is half the top speed reached during deceleration. The average speed during the acceleration stage is therefore higher than the average speed during the deceleration stage, so the distance travelled while accelerating is greater than the distance travelled while decelerating.

Examiner tips

  • Use the word ‘uniform’ to describe the rate of change of speed in each stage.
  • Show that the deceleration is twice the acceleration and that the times are equal to explain the distance comparison.
  • Mention that average speed = (initial+final)/2 for each stage.
  • Keep the answer concise – 4 marks only.

Common mistakes

  • Confusing acceleration with speed; saying the train ‘speeds up’ instead of ‘velocity increases uniformly’.
  • Assuming the distances are equal because the times are equal, ignoring the different magnitudes of acceleration and deceleration.
  • Using the wrong sign for acceleration/deceleration or not stating that the deceleration is twice the acceleration.

Mark scheme (4 marks)

  1. During the acceleration stage, the train's speed increases at a steady/uniform rate (velocity increases uniformly from zero).
  2. During the constant speed stage, the train travels at the same speed / velocity does not change / acceleration is zero.
  3. During the deceleration stage, the train's speed decreases at a steady/uniform rate (to rest).
  4. The distance during acceleration is greater than the distance during deceleration, because the average speed during acceleration (starting from zero and reaching top speed) is the same as during deceleration (starting from top speed and reaching zero) — but since deceleration is twice as large, the deceleration stage takes half the time to reach zero from top speed if time were equal, meaning if time IS the same, the top speed reached during acceleration is half that during deceleration, so the train covers less distance during deceleration / OR: same time but deceleration is twice as large means the final speed drops twice as fast, so average speed over deceleration is lower, covering less distance.

Key terms in this question

deceleration · constant speed · distance

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