4PM1/01R Jan 2020, Question 8: Motion with Calculus
Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 8 marks
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The question, in short
Particle P moves on the positive x-axis with v = 3 + 5t − 2t², starting at x = 5 when t = 0. Find the maximum x and justify it is a maximum.
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Solution · 8 marks
- 1x = ∫ (3 + 5t − 2t²) dt = 3t + 5t²/2 − 2t³/3 + c
- 2t = 0, x = 5 gives c = 5
- 3dx/dt = 0: 2t² − 5t − 3 = 0, (2t + 1)(t − 3) = 0, t = 3
- 4x = 5 + 9 + 45/2 − 18 = 37/2
- 5d²x/dt² = 5 − 4t = −7 < 0 at t = 3, so maximum
AnswerMaximum x = 37/2 (18.5), at t = 3; d²x/dt² = −7 < 0 so maximum
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