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4PM1/01R Jan 2023, Question 7: Motion with Calculus

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 11 marks

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The question, in short

Particle P moves on the x-axis with velocity v = t² − 10t + 28 for t ≥ 0. Use this and the given position at t = 3 to find velocity, distance, acceleration and minimum speed.

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

Part (a) · 1 mark

Calculate the velocity of P at t = 1 using the given velocity expression.

  1. 1v = 1² − 10(1) + 28
  2. 2v = 19 m/s

Answer19 m/s

Part (b) · 5 marks

Given P is 24 m from the origin at t = 3, find its distance from the origin at t = 5.

  1. 1s = ∫ (t² − 10t + 28) dt = t³/3 − 5t² + 28t + c
  2. 2t = 3, s = 24: 24 = 9 − 45 + 84 + c → c = −24
  3. 3s = t³/3 − 5t² + 28t − 24
  4. 4t = 5: s = 125/3 − 125 + 140 − 24 = 98/3 m

Answer98/3 m (32.7 m)

Part (c) · 2 marks

Find the acceleration of P at t = 9.

  1. 1a = dv/dt = 2t − 10
  2. 2t = 9: a = 18 − 10 = 8 m/s²

Answer8 m/s²

Part (d) · 3 marks

Show P is never instantaneously at rest, then find the smallest magnitude of its velocity.

  1. 1v = t² − 10t + 28 = (t − 5)² + 3
  2. 2(t − 5)² ≥ 0 so v ≥ 3 > 0: P is never at rest
  3. 3(or discriminant (−10)² − 4(1)(28) = −12 < 0: no real roots)
  4. 4Least magnitude of velocity = 3 m/s (at t = 5)

Answer(i) v = (t − 5)² + 3 ≥ 3 > 0, so P is never at rest; (ii) 3 m/s

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