A-LevelMathematicsMechanics9709/41 · May/June 2020 · Q611 Marks
A particle moves in a straight line AB. The velocity v m s⁻¹ of the particle t s after leaving A is given by v = k(t² – 10t + 21), where k is a constant. The displacement of the particle from A, in the direction towards B, is 2.85 m when t = 3 and is 2.4 m when t = 6.
- (a)Find the value of k. Hence find an expression, in terms of t, for the displacement of the particle from A.[7]
- (b)Find the displacement of the particle from A when its velocity is a minimum.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.
- [M1] ∫ k(t² - 10t + 21) dt
- [A1] s = k(t³/3 - 5t² + 21t) + C
- [M1] 2.85 = k(3³/3 - 5×3² + 21×3) + C or 2.4 = k(6³/3 - 5×6² + 21×6) + C
- [A1] 2.85 = 27k + C, 2.4 = 18k + C (A1 for both)
- [M1] Solving for k
- [A1] k = 0.05
- [A1] s = 0.05(t³/3 - 5t² + 21t) + 1.5
- [M1] Differentiating v or completing the square for v
- [A1] a = 0.05(2t - 10)
- [M1] Min value of v is at t = 5.
- [A1] Displacement at t = 5 is 2.58 m (2.5833...)
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