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4PM1/01 Jun 2023, Question 4: Motion with Calculus

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 10 marks

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The question, in short

Particle P moves on the x-axis with velocity v = 2t² − 16t + 30 for t ≥ 0. Find the acceleration at t = 5, then the exact distance MN between the two points where P is instantaneously at rest.

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Part (a) · 2 marks

Find the acceleration of P in m/s² at the instant t = 5.

  1. 1a = dv/dt = 4t − 16
  2. 2When t = 5: a = 20 − 16 = 4 m/s²

Answer4 m/s²

Part (b) · 8 marks

Given P is instantaneously at rest at points M and N at times t₁ and t₂ with t₂ > t₁, find the exact distance MN.

  1. 1v = 0: 2t² − 16t + 30 = 0, so t² − 8t + 15 = 0
  2. 2(t − 3)(t − 5) = 0, so t₁ = 3, t₂ = 5
  3. 3Displacement = ∫ from 3 to 5 of (2t² − 16t + 30) dt = [2t³/3 − 8t² + 30t] from 3 to 5
  4. 4= (250/3 − 200 + 150) − (18 − 72 + 90) = 100/3 − 36 = −8/3
  5. 5Distance MN = 8/3 m

AnswerMN = 8/3 m

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