A particle P moving in a straight line starts from rest at a point O and comes to rest 16 s later. At time t s after leaving O, the acceleration a m s⁻² of P is given by a = 6 + 4t 0 ≤ t < 2, a = 14 2 < t < 4, a = 16 - 2t 4 ≤ t ≤ 16. There is no sudden change in velocity at any instant.

Mark scheme

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

1. [M1] Attempt to integrate a in Stage 1 or Stage 2 or in Stage 2 for use of v=u+at
2. v = 6t + 2t² [+c] or v = 14t [+c]
3. [A1] Velocity in Stage 1 and Stage 2 correct including correct constant Find v at t = 2 and use v=u+14t to find v at t=4
4. v=6t+2t² and v=14t-8
5. v(t = 2) = 20
6. v(t = 4) = 20+14×2=48
7. [*M1] Attempt to integrate a in Stage 3.
8. v = 16t-t² [+c]
9. [DM1] Attempt to solve a relevant 3-term quadratic equation which comes from their 2 term v from Stage 3 equated to 55 and finding two values of t
10. 55=16t-t²
11. [A1] t = 5 and t = 11 only Allow only if c = 0 has been shown correctly.
12. [B1] Positive quadratic for 0<t< 2 through (0,0) joining to the bottom of the given line or Negative quadratic for 4 ≤ t ≤ 16 going through the point (16,0) and joining the top of the given line
13. [B1] All correct with correct gradients (approx) Negative quadratic must have a maximum. There must be no point of inflexion particularly near t=16. Ignore any curve drawn outside 0 ≤ t ≤ 16.
14. [M1] Attempt to integrate their v.
15. s=[(16t-t²)dt = 8t²-1/3 t³ (+c)
16. [A1] Correct integral and the correct limits used correctly to find an unsimplified expression for the distance from t=8 to t=16 only.
17. s = [8t² - 1/3 t³] from 8 to 16
18. s = [2048 - 1365 1/3] - [512 - 170 2/3]
19. [B1] s = 341 1/3 Allow s = 341 to 3s.f. If no integration seen (calculator used) allow B1 (max 1 out of 3 marks)