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IGCSE Maths · Edexcel 4XMA1 · Unit 2 Higher · Calculus

Calculus: Kinematics and Practical Problems — Worked Solutions

Full worked solutions to all 12 questions (39 marks), set out the way an Edexcel examiner wants to see them.

Section A — Warm-up (grades 6–7)
  1. (2 marks)s = 2t³ − 5t + 1. Find the velocity when t = 2.
    Solution
    1. v = 6t² − 5.
    2. At t = 2: 24 − 5 = 19.
    19 m/s
  2. (2 marks)The velocity of a particle is v = 4t² − 3t m/s. Find its acceleration when t = 3.
    Solution
    1. a = 8t − 3.
    2. At t = 3: 24 − 3 = 21.
    21 m/s²
  3. (3 marks)s = t³ − 6t² + 9t. Find the times when the particle is at rest.
    Solution
    1. v = 3t² − 12t + 9 = 0.
    2. t² − 4t + 3 = 0 → (t − 1)(t − 3) = 0.
    t = 1 s and t = 3 s
  4. (3 marks)The height of a ball after t seconds is h = 20t − 5t² metres. Find the maximum height.
    Solution
    1. dh/dt = 20 − 10t = 0 → t = 2.
    2. h = 40 − 20 = 20. (∩-shaped, so this is a maximum.)
    20 m
Section B — Core (grades 7–8)
  1. (3 marks)s = t³ − 3t² + 4, for t > 0. Find the acceleration at the moment when the particle is instantaneously at rest.
    Solution
    1. v = 3t² − 6t = 3t(t − 2) = 0 → t = 2 (t > 0).
    2. a = 6t − 6. At t = 2: a = 6.
    6 m/s²
  2. (3 marks)v = 3t² − 18t + 24. Find the minimum velocity.
    Solution
    1. dv/dt = 6t − 18 = 0 → t = 3.
    2. v = 27 − 54 + 24 = −3. (Positive t² term, ∪ shape, so it is a minimum.)
    −3 m/s
  3. (3 marks)A rectangle has perimeter 36 cm and width x cm.
    (a) Show that its area is A = 18x − x².
    (b) Use calculus to find the maximum area.
    Solution
    1. (a) Length = (36 − 2x) ÷ 2 = 18 − x, so A = x(18 − x) = 18x − x².
    2. (b) dA/dx = 18 − 2x = 0 → x = 9.
    3. A = 9 × 9 = 81.
    81 cm²
  4. (4 marks)A farmer uses 40 m of fencing to make a rectangular pen against a wall. The two sides perpendicular to the wall are each x m. Find the maximum area of the pen.
    Solution
    1. A = x(40 − 2x) = 40x − 2x².
    2. dA/dx = 40 − 4x = 0 → x = 10.
    3. A = 10 × 20 = 200.
    200 m²
Section C — Stretch (grades 8–9)
  1. (4 marks)The cost of a journey is C = 2x + 50x dollars, where x > 0. Find the value of x that gives the minimum cost, and the minimum cost.
    Solution
    1. C = 2x + 50x⁻¹, so dC/dx = 2 − 50x⁻².
    2. 2 − 50x² = 0 → x² = 25 → x = 5.
    3. C = 10 + 10 = 20.
    4. At x = 4: dC/dx = 2 − 50/16 < 0; at x = 6: 2 − 50/36 > 0. − then +, so it is a minimum.
    x = 5; minimum cost $20
  2. (4 marks)An open box is made from a square sheet of card 24 cm by 24 cm by cutting squares of side x cm from each corner.
    (a) Show that V = x(24 − 2x)².
    (b) Find the maximum volume.
    Solution
    1. (a) The base is (24 − 2x) by (24 − 2x) and the height is x.
    2. (b) V = 576x − 96x² + 4x³, so dV/dx = 576 − 192x + 12x² = 12(x − 4)(x − 12).
    3. x = 12 gives no box, so x = 4.
    4. V = 4 × 16² = 1024.
    1024 cm³
  3. (4 marks)A particle has displacement s = t³ − 9t² + 24t metres. Find the total distance it travels in the first 4 seconds.
    Solution
    1. v = 3t² − 18t + 24 = 3(t − 2)(t − 4), so it turns round at t = 2.
    2. s(0) = 0, s(2) = 8 − 36 + 48 = 20, s(4) = 64 − 144 + 96 = 16.
    3. Distance = 20 (out) + 4 (back) = 24.
    24 m
  4. (4 marks)Find the coordinates of the stationary point of y = x³ − 12x + 5 where x > 0, and show that it is a minimum.
    Solution
    1. dy/dx = 3x² − 12 = 0 → x = ±2; take x = 2.
    2. y = 8 − 24 + 5 = −11.
    3. At x = 1: dy/dx = −9; at x = 3: dy/dx = 15. − then +, so it is a minimum.
    (2, −11), minimum
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