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0606/22 Feb/Mar 2026, Question 10: Motion with Calculus

Cambridge IGCSE Additional Mathematics 0606 · Paper 22 (Calculator) · Calculator · 10 marks

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The question, in short

Particle on positive x-axis with acceleration 2t − 9, velocity 4 at t = 7, displacement 150 at t = 12. Find zero-velocity times, displacement expression, and distance in first 6 s.

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

Part (a) · 4 marks

Find the times when the particle's velocity equals 0.

  1. 1v = ∫ (2t − 9) dt = t² − 9t + c
  2. 2At t = 7: 4 = 49 − 63 + c, so c = 18
  3. 3v = 0: t² − 9t + 18 = 0
  4. 4(t − 3)(t − 6) = 0, t = 3 and t = 6

Answert = 3 and t = 6

Part (bi) · 3 marks

Find an expression for the particle's displacement at time t.

  1. 1s = ∫ (t² − 9t + 18) dt = t³/3 − 9t²/2 + 18t + k
  2. 2At t = 12: 150 = 576 − 648 + 216 + k
  3. 3k = 6
  4. 4s = t³/3 − 9t²/2 + 18t + 6

Answers = t³/3 − 9t²/2 + 18t + 6

Part (bii) · 3 marks

Find the distance travelled by the particle during the first 6 seconds.

  1. 1The particle stops at t = 3, so split the interval at t = 3 and t = 6
  2. 2s(0) = 6, s(3) = 28.5, s(6) = 24
  3. 3s₃ − s₀ = 22.5
  4. 4s₆ − s₃ = −4.5
  5. 5Distance = 22.5 + 4.5 = 27.0 m

Answer27.0 m

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