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0606/21 May/Jun 2025, Question 10: Motion with Calculus

Cambridge IGCSE Additional Mathematics 0606 · Paper 21 (Calculator) · Calculator · 11 marks

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The question, in short

Particle P moves in a line with acceleration defined piecewise for 0 ≤ t ≤ 4 and 4 ≤ t ≤ 10; given velocity and displacement at t = 3, find velocity and displacement at t = 4 and displacement at t = 10.

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Part (ai) · 3 marks

Find the velocity of P at t = 4 using the acceleration for 0 ≤ t ≤ 4 and the velocity at t = 3.

  1. 1v = ∫ (t² − 2) dt = t³/3 − 2t + c
  2. 2At t = 3: 9 − 6 + c = −1/3, so c = −10/3
  3. 3v = t³/3 − 2t − 10/3
  4. 4At t = 4: v = 64/3 − 8 − 10/3 = 10

Answerv = 10 m/s

Part (aii) · 3 marks

Find the displacement of P from O at t = 4 using the velocity and the displacement at t = 3.

  1. 1s = ∫ (t³/3 − 2t − 10/3) dt = t⁴/12 − t² − (10/3)t + d
  2. 2At t = 3: 81/12 − 9 − 10 + d = −1/4, so d = 12
  3. 3s = t⁴/12 − t² − (10/3)t + 12
  4. 4At t = 4: s = 64/3 − 16 − 40/3 + 12 = 4

Answers = 4 m

Part (b) · 5 marks

Find the displacement of P from O at t = 10 using the acceleration for 4 ≤ t ≤ 10 and values at t = 4.

  1. 1v = ∫ (19 − 5e^(8 − 2t)) dt = 19t + (5/2)e^(8 − 2t) + c
  2. 2At t = 4, v = 10: 76 + 5/2 + c = 10, so c = −137/2
  3. 3s = 19t²/2 − (5/4)e^(8 − 2t) − (137/2)t + d
  4. 4At t = 4, s = 4: 152 − 5/4 − 274 + d = 4, so d = 509/4
  5. 5At t = 10: s = 950 − (5/4)e^(−12) − 685 + 509/4 ≈ 392.2

Answers = 392 m (392.2)

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Also searched as: 0606/21/M/J/25 · 0606_s25_qp_21 · 0606 s25 21 · question 10.
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