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0580/41 May/Jun 2021, Question 2: Area Under a Speed-Time Graph

Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 8 marks

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The question, in short

Use a speed-time graph for the first 180 s of a train journey to find acceleration, show a deceleration time, complete the graph, and calculate average speed.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 1 mark

Find the train's acceleration in m/s² over the first 50 seconds from the graph.

  1. 1Acceleration = gradient = 9 ÷ 50 = 0.18 m/s²

Answer0.18 or 9/50

Part (b) · 2 marks

Given deceleration of 1944 km/h² after 180 s, show that the train takes 60 seconds to stop.

  1. 11944 km/h² = 1944 × 1000 ÷ (3600 × 3600) m/s² = 0.15 m/s²
  2. 2Time to stop = 9 ÷ 0.15 = 60 s

Answer1944 × 1000/(3600 × 3600) = 0.15 9 ÷ 0.15 = 60

Part (c) · 1 mark

Given deceleration for 60 seconds after 180 s, complete the speed-time graph.

  1. 1Draw a ruled line from (180, 9) down to the time axis at 180 + 60 = 240 s

AnswerRuled line from (180, 9) to the time axis at 240

Part (d) · 4 marks

Calculate the average speed of the train over the whole journey.

  1. 1Distance = area under graph = 1/2 × (130 + 240) × 9 = 1665 m
  2. 2Average speed = 1665 ÷ 240 = 6.9375 m/s

Answer6.9375

Stuck on a step? Read the free lesson: Area Under a Speed-Time Graph

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Also searched as: 0580/41/M/J/21 · 0580_s21_qp_41 · 0580 s21 41 · question 2.
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