A cyclist starts from rest at a fixed point O and moves in a straight line, before coming to rest k seconds later. The acceleration of the cyclist at time ts after leaving O is ams¯², where a = 2t^(1/2) - (2/5)t^(3/2) for 0<t≤k.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2022). The mark or mark type is shown in brackets.

1. [*M1] For an attempt at integration (Power of at least one term increased by 1 and the coefficient changed.)
2. [A1] v = 4/3 t^(3/2) - 4/25 t^(5/2) [+C] (Correct v)
3. [DM1] 4/3 t^(3/2) - 4/25 t^(5/2) = 0 (Equating their 2-term v to zero and attempt to solve for t or k.)
4. [A1] k = 10 (Final answer t = 10 is A0)
5. [M1] Attempt to solve a = 0 and find a value of t.
6. [A1] t = 10/3
7. [B1] Maximum speed = 4.87 ms¯¹ to 3 sf (Allow maximum speed as 8√30/9)
8. [*M1] For an attempt at integration of their v (Power of at least one term increased by 1 and the coefficient changed.)
9. [A1] s = 8/15 t^(5/2) - 8/175 t^(7/2) [+C] (Correct s)
10. [DM1] Substitute their t = 10/3 and t = 10 (≠0) correctly. (Use their t=10 and their t = 10/3 correctly.)
11. [A1] Distance = 20.7 m (Distance = 20.7479...)