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0606/23 Oct/Nov 2025, Question 10: Motion with Calculus

Cambridge IGCSE Additional Mathematics 0606 · Paper 23 (Calculator) · Calculator · 11 marks

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The question, in short

Given s = (10t + 100)/√(2t² + 100) for a particle moving from A, use the displacement–time graph (0 ≤ t ≤ 30) to find when s is greatest, the distance travelled by the return time T, and T itself.

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

Part (a) · 6 marks

Find the time t at which the displacement s reaches its maximum value.

  1. 1d/dt √(2t² + 100) = ½(2t² + 100)^(−1/2) × 4t = 2t/√(2t² + 100)
  2. 2ds/dt = (10√(2t² + 100) − (10t + 100) × 2t/√(2t² + 100))/(2t² + 100)
  3. 3ds/dt = 0: 10(2t² + 100) − 2t(10t + 100) = 0
  4. 420t² + 1000 − 20t² − 200t = 0
  5. 51000 − 200t = 0, so t = 5

Answert = 5

Part (bi) · 2 marks

Given the particle returns to its starting point at time T, find the total distance travelled during the first T seconds.

  1. 1At t = 0, s = 100/10 = 10; at t = 5, s = 150/√150 = √150 (maximum)
  2. 2The particle goes from s = 10 up to √150 and back to s = 10
  3. 3Distance = 2(√150 − 10) ≈ 4.49 m

Answer2(√150 − 10) or 4.49 m

Part (bii) · 3 marks

Use algebra to determine the value of T when the particle is back at its starting point.

  1. 1(10t + 100)/√(2t² + 100) = 10
  2. 2(10t + 100)² = 100(2t² + 100)
  3. 3100t² + 2000t + 10 000 = 200t² + 10 000
  4. 4100t² = 2000t, so t = 0 or t = 20
  5. 5T = 20

AnswerT = 20

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