A sprinter runs 100 m along a straight track in 12.5 s and then immediately turns around and jogs 40 m back towards the starting line in 20 s. Explain why the sprinter's average speed for the whole journey is different from the magnitude of the sprinter's average velocity for the whole journey.

Cambridge International IGCSE Physics (0625) — 1.2 Motion · Explain · 4 marks · View as Markdown

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

Model answer (4 marks)

Average speed is the total distance travelled divided by the total time, so it is 140 m ÷ 32.5 s = 4.31 m s⁻¹. Average velocity is the displacement (the straight‑line change in position) divided by the total time, so it is 60 m ÷ 32.5 s = 1.85 m s⁻¹. Speed is a scalar and does not depend on direction, whereas velocity is a vector and its magnitude depends on the net displacement. Because the total distance (140 m) is greater than the magnitude of the displacement (60 m) while the time is the same, the average speed is greater than the magnitude of the average velocity.

Examiner tips

  • Show the two formulas separately, then plug in the numbers; keep units
  • Explain the scalar vs vector nature of speed and velocity
  • State why the distance is larger than the displacement
  • Use the word ‘magnitude’ when referring to average velocity

Common mistakes

  • Using 100 m as the total distance instead of 140 m
  • Confusing displacement with total distance
  • Forgetting to divide by the same total time for both calculations

Mark scheme (4 marks)

  1. Average speed uses total distance travelled (140 m)
  2. Average velocity uses displacement (60 m), which is the overall change in position / distance from start to finish point
  3. Speed is a scalar (distance ÷ time) so direction is not considered, whereas velocity is a vector (displacement ÷ time) so direction is considered
  4. Because total distance (140 m) is greater than the magnitude of displacement (60 m), and the total time is the same for both calculations, average speed is greater than the magnitude of average velocity

Key terms in this question

average speed · average velocity

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