A cyclist starts from rest at a point A and travels along a straight road AB, coming to rest at B. The displacement of the cyclist from A at time t s after the start is s m, where s = 0.004(75t² – t³).

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

1. *M1 For differentiation of s. v = 0.004(150t – 3t²) [= 0.6t – 0.012t²]
2. A1 v = 0 when t = 50. At t = 50, s = 0.004(75 × 50² – 50³) = 0.3 × 50² – 0.004 × 50³
3. DM1 Solve v = 0 for t and substitute this value into s.
4. A1 AG Distance AB = 250 m
5. *M1 Attempt to determine stationary points for v by differentiation or by use of symmetry [a = 0.004(150 – 6t) = 0.6 – 0.024t] or using symmetry attempt to find the mid-point between t = 0 and their t value at v=0. If symmetry used then an attempt to find the required mid-point must be seen.
6. DM1 Attempt to solve a = 0 or use symmetry to find the relevant t value.
7. A1 Maximum velocity = 7.5 ms⁻¹
8. Alternative method for question 4(b)
9. M1* Attempt to complete the square for their velocity as far as k[(t-a)²-a²]
10. DM1 Or select the 0.012×25² term as the maximum velocity.
11. A1 Maximum = [0.012×625=]7.5 ms⁻¹