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0606/22 Feb/Mar 2025, Question 9: Motion with Calculus

Cambridge IGCSE Additional Mathematics 0606 · Paper 22 (Calculator) · Calculator · 11 marks

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The question, in short

A particle P has displacement s = 5 ln(2t + 1) − 5t metres after t seconds from fixed point O. Use this to find a time, a distance, and acceleration values.

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

Part (a) · 4 marks

Determine the time t at which the particle's velocity is zero, so it is momentarily stationary.

  1. 1v = ds/dt = 10/(2t + 1) − 5
  2. 2v = 0: 10/(2t + 1) = 5
  3. 32t + 1 = 2
  4. 4t = 1/2

Answert = 1/2

Part (b) · 4 marks

Calculate how far P moves over the interval from t = 0 to t = 2.

  1. 1P changes direction at t = 1/2
  2. 2t = 1/2: s = 5 ln 2 − 2.5 = 0.9657
  3. 3t = 2: s = 5 ln 5 − 10 = −1.9528
  4. 4Distance = 2(0.9657) + 1.9528 = 3.88 m

Answer3.88

Part (c) · 2 marks

Derive a formula giving P's acceleration as a function of time t.

  1. 1v = 10(2t + 1)⁻¹ − 5
  2. 2a = dv/dt = −10 × 2(2t + 1)⁻² = −20/(2t + 1)²

Answera = −20/(2t + 1)²

Part (d) · 1 mark

Evaluate the acceleration at the instant t = 4.5.

  1. 1a = −20/(2(4.5) + 1)² = −20/100 = −0.2 m/s²

Answer−0.2

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