GCSE Physics  ›  P5.7 Motion

Motion

Free GCSE Physics practice questions on Motion. Aligned with the UK Department for Education GCSE subject content — works for any UK GCSE exam board. Sample questions below with detailed mark schemes. Sign up to practise the full set with spaced repetition.

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Describe — 2 marks

A cyclist is travelling along a straight road. She accelerates from rest for 5 seconds, reaching a speed of 8 m/s. She then travels at this constant speed for 10 seconds before applying the brakes and decelerating uniformly to a stop over 4 seconds.

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  • (a) The cyclist is accelerating / increasing velocity uniformly / at a constant rate from 0 m/s to 8 m/s
  • (a) Accept: The cyclist's speed increases / she starts from rest and speeds up
  • (b) The velocity decreases uniformly / at a constant rate from 8 m/s to 0 m/s
  • (b) Accept: The cyclist decelerates / slows down at a constant rate / velocity changes by the same amount each second

Calculate — 2 marks

A cyclist travels in a straight line along a flat road. She accelerates uniformly from rest and reaches a velocity of 8 m/s after 4 seconds.

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  • (a) Correct application of a = (v - u) / t with correct substitution: a = (8 - 0) / 4 = 2 m/s² OR equivalent correct method. Award 1 mark for correct answer of 2 m/s².
  • (b) Correct application of s = v × t with correct substitution: s = 8 × 10 = 80 m OR equivalent correct method. Award 1 mark for correct answer of 80 m.

Compare — 5 marks

A cyclist and a runner both travel along the same 100 m straight path. The cyclist accelerates uniformly from rest and reaches 8 m/s after 10 seconds, then maintains this constant velocity. The runner accelerates uniformly from rest and reaches 6 m/s after 12 seconds, then maintains this constant velocity.

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  • (a) Cyclist acceleration = 8 ÷ 10 = 0.8 m/s² (1 mark)
  • (a) Runner acceleration = 6 ÷ 12 = 0.5 m/s², so the cyclist has a greater acceleration by 0.3 m/s² (1 mark)
  • (b) Cyclist distance during acceleration = ½ × 0.8 × 10² = 40 m (1 mark)
  • (b) Runner distance during acceleration = ½ × 0.5 × 12² = 36 m, so the cyclist travels 4 m further during acceleration (1 mark)
  • (c) The cyclist reaches constant velocity sooner and has already covered more distance (40 m vs 36 m), so the cyclist completes the 100 m journey in less time / achieves a higher speed allowing them to cover the remaining distance faster (1 mark)

Define — 5 marks

A traffic engineer is analysing the motion of vehicles on a motorway. She needs to understand the difference between how fast a car is travelling at any given instant and how far it travels over a longer journey. The data from speed cameras and GPS tracking systems provides different types of information about vehicle motion.

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  • (a) The speed of an object at a particular / specific moment in time (1 mark)
  • (a) Measured over a very small / infinitesimal time interval (1 mark)
  • (b) Average speed is the total distance travelled divided by the total time taken (1 mark)
  • (b) Instantaneous speed is the speed at one moment, whereas average speed is calculated over a longer time period / journey (1 mark)
  • (c) The instantaneous speed (100 km/h) is the speed at one particular moment, but the average speed over 3 hours is 80 km/h (240÷3), so the car must have travelled faster at some points and slower at others during the journey (1 mark)
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