A-LevelMathematicsMechanics9709/43 · May/June 2020 · Q69 Marks
A particle travels in a straight line PQ. The velocity of the particle t s after leaving P is v m s⁻¹, where v = 4.5 + 4t – 0.5t².
- (a)Find the velocity of the particle at the instant when its acceleration is zero.[3]
- (b)The particle comes to instantaneous rest at Q. Find the distance PQ.[6]
Mark scheme
Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.
- [M1] a = 4 – t (for differentiation)
- [A1] When a = 0, t = 4
- [A1] At t = 4, v = 12.5
- [M1] Velocity = 0 when 4.5 + 4t – 0.5t² = 0
- [A1] t = 9 (reject t = -1)
- [M1] ∫(4.5 + 4t – 0.5t²)dt
- [A1] 4.5t + 2t² – 1/6t³ [+c]
- [M1] Apply limits (0 and 9)
- [A1] Distance = 81 m
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