A particle moves in a straight line. It starts from rest, at time t = 0, and accelerates at 0.6tms⁻² for 4s, reaching a speed of Ums⁻¹. The particle then travels at Vms⁻¹ for 11s, and finally slows down, with constant deceleration, stopping after a further 5 s.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.

1. [B1] AG Must see 0.6t^2/2 or 0.3t^2 and 4 must be actually shown substituted. Merely stating t = 4 is not enough to score this mark. Do not allow use of s = ut + 1/2at^2 which leads to 4.8 if a = 0.6 is used.
2. [B1] Note: the grid is for reference – not shown in QP. Their graph does not need to be to scale.
3. [B1] The points should be specified somehow, but for an accurate sketch allow a line just below 5 without specifying 4.8.
4. [B1] The points should be specified somehow, but for an accurate sketch allow a line just below 5 without specifying 4.8. Allow all 3 marks if using V instead of 4.8. ISW any extra out of the range for t of 0 to 20. If no marks scored then SC B1 for a correct shaped graph with no numbers. If using a value of v≠4.8, allow SC B1 for the first section correct and SC B1 for the second and third both correct.
5. [M1] For calculation (0-4.8)/(20-15) oe [=-0.96] Allow +0.96.
6. [A1] Oe e.g. Allow [v=]4.8-0.96(t-15).
7. [*M1] Attempt to integrate their v from part (a) provided this came from integration, but allow a restart here. The power of t must increase by 1 with a change of coefficient. Use of s = vt scores M0. No need for limits. If no integration seen allow SCM1 for answer of 6.4 in place of M1M1.
8. [DM1] Correct use of correct limit(s) (expect 6.4).
9. [B1] Both correct and added. May be done in one go using a trapezium (11+16)×4.8/2. Could do the last stage by integration. Maximum B1 for final answer 74.4 from thinking the first section is also straight.
10. [A1]