A-LevelMathematicsIntegration9709/43 · Oct/Nov 2017 · Q58 Marks
A particle starts from a fixed origin with velocity 0.4 m s⁻¹ and moves in a straight line. The acceleration a m s⁻² of the particle t s after it leaves the origin is given by a = k(3t² – 12t + 2), where k is a constant. When t = 1, the velocity of P is 0.1 m s⁻¹.
- (i)Show that the value of k is 0.1.[5]
- (ii)Find an expression for the displacement of the particle from the origin in terms of t.[2]
- (iii)Hence verify that the particle is again at the origin at t = 2.[1]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.
- v = ∫ k(3t² – 12t + 2) dt *M1 Use of v = ∫ a dt
- = k(3t³/3 – 12t²/2 + 2t) + C
- v=k(t³-6t²+2t)+C A1 Condone C missing
- C = 0.4 B1
- 0.1 = k(1-6+2) +0.4 [-0.3 = -3k] DM1 Substitutes t = 1, v = 0.1
- k = 0.1 A1 AG
- s = ∫ 0.1(t³ – 6t² + 2t) + 0.4 dt M1 Use of s = ∫ v dt
- = 0.1(t⁴/4 – 6t³/3 + 2t²/2) + 0.4t + C
- s = 0.025t⁴ – 0.2t³ + 0.1t² + 0.4t A1 C = 0 seen or implied
- Substitutes t = 2 to show s = 0 B1 AG
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