O-LevelAdditional MathematicsDifferentiation and integration4037/12 · Oct/Nov 2016 · Q1012 Marks

The diagram shows part of the velocity-time graph for a particle, moving at vms⁻¹ in a straight line, ts after passing through a fixed point. The particle travels at U ms⁻¹ for 20s and then decelerates uniformly for 10s to a velocity of U/2 ms⁻¹. In this 30s interval, the particle travels 165 m. [Figure 10.1]

(a)The diagram shows part of the velocity-time graph for a particle, moving at vms⁻¹ in a straight line, ts after passing through a fixed point. The particle travels at U ms⁻¹ for 20s and then decelerates uniformly for 10s to a velocity of U/2 ms⁻¹. In this 30s interval, the particle travels 165 m. [Figure 10.1]
(a)(i)Find the value of U.[3]
(a)(ii)Find the acceleration of the particle between t = 20 and t = 30.[2]
(b)A particle P travels in a straight line such that, ts after passing through a fixed point O, its velocity, v ms⁻¹, is given by v = (e^(t/8) - 4)³.
(b)(i)Find the speed of P at O.[1]
(b)(ii)Find the value of t for which P is instantaneously at rest.[2]
(b)(iii)Find the acceleration of P when t = 1.[4]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2016). The mark or mark type is shown in brackets.

[1] (a)(i) Area under graph = 20U + 1/2 (U + U/2) * 10 M1 for realising that area under the graph is needed and attempt to find an area
[1] (a)(i) 20U + 1/2 (3U/2) * 10 = 165 DM1 for equating their area to 165 and attempt to solve
[1] (a)(i) U = 6 A1
[2] (a)(ii) Acceleration = gradient = (U/2 - U) / (30 - 20) = (-U/2) / 10 = -U/20. With U=6, acceleration = -6/20 = -0.3 ms⁻² M1 for use of the gradient, must be negative, A1 for correct value
[1] (b)(i) At O, t=0. v = (e^(0/8) - 4)³ = (1 - 4)³ = (-3)³ = -27. Speed = |-27| = 27 B1
[1] (b)(ii) At rest, v=0. (e^(t/8) - 4)³ = 0 => e^(t/8) - 4 = 0 => e^(t/8) = 4 M1 for a correct attempt to solve e^(t/8) = 4
[1] (b)(ii) t² = 8 ln 4 => t = 3.33 or better A1
[1] (b)(iii) dv/dt = 3(e^(t/8) - 4)² * (d/dt (e^(t/8) - 4)) = 3(e^(t/8) - 4)² * (1/8 e^(t/8)) M1 for a correct attempt to differentiate using the chain rule
[1] (b)(iii) dv/dt = (3/8)e^(t/8)(e^(t/8) - 4)² A1
[1] (b)(iii) At t=1, a = (3/8)e^(1/8)(e^(1/8) - 4)² M1 for use of t = 1 in their acceleration
[1] (b)(iii) a ≈ 6.98 A1

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