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0606/23 May/Jun 2026, Question 5: Area Under a Speed-Time Graph

Cambridge IGCSE Additional Mathematics 0606 · Paper 23 (Calculator) · Calculator · 8 marks

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The question, in short

Particle starts from rest, accelerates to V at time T, moves at V until 3T, then decelerates to rest at t = 60. Sketch the velocity-time graph and find T and V.

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

Part (a) · 2 marks

Sketch a velocity-time graph on the given axes showing rest at t = 0, velocity V at t = T, constant V until t = 3T, and rest at t = 60.

  1. 1Constant acceleration from rest: straight line from O to (T, V)
  2. 2Constant velocity: horizontal line from (T, V) to (3T, V)
  3. 3Uniform deceleration: straight line from (3T, V) to (60, 0)
  4. 4Label T, 3T, 60 on the time axis and V on the velocity axis

AnswerTrapezium: straight line from O up to (T, V), horizontal line at V from T to 3T, straight line down to (60, 0); T, 3T, 60 and V marked

Part (b) · 6 marks

Given acceleration 1.5 during the first stage and total distance 1104, find the values of T and V.

  1. 1Acceleration = V/T = 1.5 ⇒ V = 1.5T
  2. 2Distance = area of trapezium = (1/2)V(60 + (3T − T)) = V(30 + T)
  3. 31.5T(30 + T) = 1104
  4. 41.5T² + 45T − 1104 = 0 ⇒ T² + 30T − 736 = 0
  5. 5(T − 16)(T + 46) = 0 ⇒ T = 16 (T > 0)
  6. 6V = 1.5 × 16 = 24

AnswerT = 16, V = 24

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