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4PM1/01 Jun 2025, Question 1: Motion with Calculus

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 7 marks

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The question, in short

A particle P travels along a straight line with velocity v = 3t² − 9t + 7 at time t seconds. Show it never stops, find its acceleration at t = 4, and use algebra to find the distance covered for 0 ≤ t ≤ 5.

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Part (a) · 2 marks

Demonstrate that the velocity expression cannot equal zero for any time t.

  1. 1P is at rest when 3t² − 9t + 7 = 0
  2. 2b² − 4ac = (−9)² − 4 × 3 × 7 = −3
  3. 3−3 < 0, so there are no real solutions: P never comes to rest

AnswerDiscriminant = −3 < 0, so v = 0 has no real solutions and P never comes to rest

Part (b) · 2 marks

Calculate the acceleration of P at the instant t = 4, in m/s².

  1. 1a = dv/dt = 6t − 9
  2. 2t = 4: a = 24 − 9 = 15

Answer15 m/s²

Part (c) · 3 marks

By algebraic means, determine the total distance P covers between t = 0 and t = 5, in m.

  1. 1v is never 0, so distance = ∫ from 0 to 5 of v dt
  2. 2= [t³ − (9/2)t² + 7t] from 0 to 5
  3. 3= 125 − 112.5 + 35 = 95/2

Answer95/2 m

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