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0606/13 May/Jun 2025, Question 11: Motion with Calculus

Cambridge IGCSE Additional Mathematics 0606 · Paper 13 (Non-calculator) · Non-calculator · 10 marks

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The question, in short

Particle P moves along a line through fixed point O, with displacement s given by two formulas for 0 ≤ t ≤ 3 and 3 ≤ t ≤ k; velocity positive and acceleration negative in the second interval. Sketch velocity against time and find k given total distance 57 m.

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

Part (a) · 4 marks

Given maximum velocity at t = 2, draw the velocity–time graph for 0 to k on the blank axes.

  1. 1For 0 ≤ t ≤ 3: v = ds/dt = 1 + 12t − 3t²
  2. 2v(0) = 1, v(2) = 13 (maximum), v(3) = 10
  3. 3For 3 ≤ t ≤ k: v = 12 − (2/3)t, a straight line, v(3) = 10
  4. 4Velocity stays positive, so the line ends above the t-axis at t = k

AnswerFor 0 ≤ t ≤ 3: v = 1 + 12t − 3t², a curve starting at v = 1, rising to a maximum (13) at t = 2, then falling to v = 10 at t = 3. For 3 ≤ t ≤ k: v = 12 − (2/3)t, a straight line with negative gradient from (3, 10) down to t = k, ending above the t-axis

Part (b) · 6 marks

Using total distance 57 m and equal distance and displacement at t = 3, determine the constant k.

  1. 1When t = 3, s = 3 + 54 − 27 = 30, so distance for 0 ≤ t ≤ 3 is 30 m
  2. 2Distance for 3 ≤ t ≤ k = 57 − 30 = 27 m
  3. 3v > 0, so s(k) − s(3) = 27: 12k − k²/3 − 3 − 30 = 27
  4. 4−k²/3 + 12k − 60 = 0, i.e. k² − 36k + 180 = 0
  5. 5(k − 6)(k − 30) = 0, k = 6 or k = 30
  6. 6k = 30 gives negative velocity (12 − 20 < 0), so k = 6

Answerk = 6

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