A particle P moves on the x-axis from the origin O with an initial velocity of −20ms⁻¹. The acceleration a ms⁻² at time ts after leaving O is given by a = 12 – 2t.

Mark scheme

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

1. [M1] Attempt to integrate 12-2t For integration, the power of t must increase by 1 in at least 1 term with a change of coefficient in the same term. No +c required for this mark. s = ∫vt is M0.
2. [A1] v = 12t - t² (+c) = 12t - t² (+c) No +c required for this mark. Allow unsimplified.
3. [B1] Use boundary conditions to get c = -20
4. [B1] Solve 12t - t² – 20 = 0 to get t = 2 and t = 10
5. [B1] Correct graph inverted quadratic starting at (0,-20) and ending at (12,-20) Ignore anything outside 0 ≤ t ≤ 12. t = 2 and t = 10 need not be shown.
6. [*M1] Attempt to integrate their 12t – t² – 20 Integrating their 2 or 3 term expression for v from (a) which has come from integration. For integration, the power of t must increase by 1 in at least 1 term with a change of coefficient in the same term.
7. [A1ft] s = 6t² - (1/3)t³ - 20t (+d) ft their +c ≠ 0. Allow unsimplified.
8. [DM1] Attempt to evaluate their [6t² - (1/3)t³ - 20t] for any of t = 0 to t = their 2 or t = their 2 to t = their 10 or t = their 10 to t = 12 Correct use of correct limits for one time interval.
9. [DM1] Attempt to evaluate their [6t² - (1/3)t³ - 20t] for all of t = 0 to t = their 2 or t = their 2 to t = their 10 or t = their 10 to t = 12 Correct use of correct limits for all 3 time intervals, ignore signs here.
10. [A1] s = |-56/3| + |200/3 - (-56/3)| + |48 - 200/3| = 368/3 = 123 m Awrt 123