A particle P moves in a straight line starting from a point O and comes to rest 14 s later. At time t s after leaving O, the velocity v m s⁻¹ of P is given by v = pt² – qt for 0 ≤ t ≤ 6, v = 63 – 4.5t for 6 ≤ t ≤ 14, where p and q are positive constants. The acceleration of P is zero when t = 2.

Mark scheme

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

1. [M1] Attempt to differentiate v. a = 2pt - q
2. [DM1] For attempting to set up 2 equations using a = 0 at t = 2 and matching the velocities at t = 6 and solve for p or q. 36p - 6q = 36. 4p - q = 0
3. [A1] p = 3, q = 12. Both correct.
4. [B1] Correct quadratic from t = 0 to t = 6. No labelling necessary for this mark.
5. [B1] Correct straight line from 6 to 14. Must join and no labelling needed.
6. [B1] All correct, labelled at (4, 0), (6, 36) and (14, 0).
7. [M1] Attempt to integrate v. Allow in terms of p and q.
8. [A1 FT] s = t³ - 6t².
9. [DM1] Using limits correctly for t = 0 to t = 6. Allow in terms of p and q. s(quadratic) = [|t³/3 - 6t²|]⁰⁶ + [|t³/3 - 6t²|]⁴⁶
10. [B1] s(triangle) = [63t - 2.25t²]⁶¹⁴ = 144 or area of triangle = 144
11. [A1] Total distance travelled in 14 s = 208 m