A-LevelMathematicsMechanics9709/43 · Oct/Nov 2019 · Q69 Marks

Particle P travels in a straight line from A to B. The velocity of P at time t s after leaving A is denoted by v m s⁻¹, where v = 0.04t³ + ct² + kt. P takes 5 s to travel from A to B and it reaches B with speed 10 m s⁻¹. The distance AB is 25 m.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

[B1] Use of t=5, v=10 (10 = 0.04×5³ + 5²c + 5k) (5c + k = 1)
[*M1] For use of s = ∫vdt (s = 0.04t⁴/4 + ct³/3 + kt²/2 (+C))
[DM1] Use of t = 0, s = 0 and t = 5, s = 25 (25 = 0.01×5⁴ + 5³c/3 + 5²k/2) (6.25 + (125/3)c + (25/2)k = 25)
[A1] (125/3)c + (25/2)k = 18.75
[M1] Solving for c or for k
[A1] c = -0.3 and k = 2.5
[M1] For use of a = dv/dt (a = 0.12t² - 0.6t + 2.5)
[M1] Uses da/dt = 0 or completes the square for a (a' = 0.24t - 0.6 = 0 → t = ... or a = 0.12(t² −5t + ...) = 0.12[(t −2.5)² +...])
[A1] AG (Minimum when t = 2.5)

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Topic: Mechanics, A-Level Mathematics (syllabus 9709). Structured question, 9 marks.
Source: A-Level Mathematics, Oct/Nov 2019 Paper 4 Variant 3, question 6.

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