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0606/21 Oct/Nov 2025, Question 8: Motion with Calculus

Cambridge IGCSE Additional Mathematics 0606 · Paper 21 (Calculator) · Calculator · 13 marks

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The question, in short

Given acceleration a = (2t − 3)² for a particle P moving through fixed point O, with v = 6 at t = 3 and displacement 4 at t = 5/2, find velocity, rest time, displacement and a small velocity change.

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Part (ai) · 3 marks

Integrate the given acceleration expression to obtain v as a function of t, using the velocity condition at t = 3.

  1. 1v = ∫ (2t − 3)² dt = (2t − 3)³/(3 × 2) + c
  2. 2When t = 3, v = 6: 27/6 + c = 6
  3. 3c = 3/2
  4. 4v = (2t − 3)³/6 + 3/2

Answerv = (2t − 3)³/6 + 3/2

Part (aii) · 3 marks

Determine the value of t at which the particle's velocity equals zero.

  1. 1At rest: (2t − 3)³/6 + 3/2 = 0
  2. 2(2t − 3)³ = −9
  3. 32t − 3 = ∛(−9) = −2.080…
  4. 4t = (3 + ∛(−9))/2 = 0.460

Answert = (3 + ∛(−9))/2 or t = 0.460

Part (aiii) · 4 marks

Using the displacement condition at t = 5/2, calculate where P is relative to O at t = 3.

  1. 1s = ∫ v dt = (2t − 3)⁴/(4 × 2 × 6) + (3/2)t + d
  2. 2When t = 5/2, s = 4: (2)⁴/48 + (3/2)(5/2) + d = 4
  3. 31/3 + 15/4 + d = 4, so d = −1/12
  4. 4s = (2t − 3)⁴/48 + (3/2)t − 1/12
  5. 5At t = 3: 81/48 + 9/2 − 1/12 = 293/48 ≈ 6.10

Answer293/48 or 6.10

Part (b) · 3 marks

Apply calculus to estimate how much v changes when t rises from 5/2 by 0.02.

  1. 1dv/dt = (2t − 3)²
  2. 2At t = 5/2: dv/dt = (5 − 3)² = 4
  3. 3δv ≈ (dv/dt) × δt = 4 × 0.02
  4. 4δv = 0.08

Answerδv = 0.08

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