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0606/13 Oct/Nov 2025, Question 4: Area Under a Speed-Time Graph

Cambridge IGCSE Additional Mathematics 0606 · Paper 13 (Non-calculator) · Non-calculator · 6 marks

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The question, in short

A velocity–time graph shows a particle speeding up for T seconds, holding a steady speed until 4 s, then slowing to rest at 8 s. Given initial acceleration 2 m s⁻² and total distance 27 m, find T and the final deceleration.

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Part (a) · 4 marks

Using the graph's shape and the 27 m total distance, work out the value of T.

  1. 1Velocity at time T is 2T
  2. 2Area of trapezium = distance: ½(8 + (4 − T)) × 2T = 27
  3. 3T(12 − T) = 27, so T² − 12T + 27 = 0
  4. 4(T − 3)(T − 9) = 0
  5. 5T = 3 (T = 9 is impossible since T < 4)

AnswerT = 3

Part (b) · 2 marks

Find the acceleration over the final 4 seconds of the motion.

  1. 1Maximum velocity = 2 × 3 = 6 m s⁻¹
  2. 2a = (0 − 6)/(8 − 4) = −1.5 m s⁻²

Answer−1.5 m s⁻²

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