(a) The diagram shows the velocity-time graph for a particle P, travelling in a straight line with velocity vms⁻¹ at a time t seconds. P accelerates at a constant rate for the first 10s of its motion, and then travels at constant velocity, 30ms⁻¹, for another 15s. P then accelerates at a constant rate for a further 10s and reaches a velocity of 60ms⁻¹. P then decelerates at a constant rate and comes to rest when t = 55. [Figure 11.1] (b) A particle Q travels in a straight line such that its velocity, `vms⁻¹`, at time `ts` after passing through a fixed point O is given by `v = 4 cos3t-4`.
- (a)(i)Find the acceleration when `t = 12`.[1]
- (a)(ii)Find the acceleration when `t = 50`.[1]
- (a)(iii)Find the total distance travelled by the particle P.[2]
- (b)(i)Find the speed of Q when `t = 5π/9`.[2]
- (b)(ii)Find the smallest positive value of t for which the acceleration of Q is zero.[3]
- (b)(iii)Find an expression for the displacement of Q from O at time t.[2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.
- [B1] 0
- [B1] -3
- [M1] For an unsimplified expression for the required area allowing at most one incorrect length
- [A1] Total distance = 1650
- [M1]
- [A1] Speed = 2
- [B1] a = -12 sin 3t
- [M1] For equating to zero and attempt to solve to obtain t, allow if in degrees
- [A1] t = π/3
- [M1] s = k sin 3t - 4t (+c)
- [A1] s = 4/3 sin 3t - 4t
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →