A particle starts from rest from a point O and moves in a straight line. The acceleration of the particle at time ts after leaving O is a m s¯², where a = kt² for 0 ≤ t ≤ 9 and where k is a constant. The velocity of the particle at t = 9 is 1.8 ms⁻¹.

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [M1] For attempt at integration. The power of t must increase by 1 with a change of coefficient. Do not penalise missing c. Use of v = at scores M0.
2. [A1] v = (k/3)t³ [+c]. Allow unsimplified e.g. v = (1/3)kt³ [+c].
3. [B1] 1.8 = kx9³ / 3. AG. Must show values substituted OE (e.g. 1.8 = 18k).
4. [M1] For attempt at integration of either. The power of t or (t −9) must increase by 1 with a change of coefficient. Use of s = vt is M0.
5. [A1] (k/12)t⁴ or (0.2/3)(t −9)³ + 1.8t. Allow unsimplified. No need for limits. Could include '+c' with either or both.
6. [M1] Correct use of limits 0, 9 or limits 9, 18. Can be implied by either answer following integration. "+c" method is as follows: s = (0.2/3)(t −9)³ + 1.8t - 243/25 or s = (0.2/3)t³ - (3.6/2)t² + 1.8t then substitute t = 18 to get 1782/25, subtract 162/25, so distance = 324/5.
7. [A1] 6.48 = (1/10) × 64.8 or 64.8 = 10 × 6.48 or 324/5 = 10 × 162/25. AG OE. Check working as can get answer from wrong working NFWW (not from wrong working).
8. [M1] For differentiation. The power of t or (t − 9) must decrease by 1 with a change of coefficient. M0 for a = -v/t.
9. [A1] Should get a = 0.4(t −9) or a=0.4t -3.6. 0.4[ms⁻²] [at t = 10]. SC B1 for 0.4 with no differentiation seen.
10. [B1] 0.3 seen (from the first phase) and state that 0.4 is final answer. No working needed. If M1A0 or M0A0 scored, then SC B1 for 0.3 without mention of the maximum acceleration.