A particle moves in a straight line. At time ts, the acceleration, ams⁻², of the particle is given by `a = 36 – 6t`. The velocity of the particle is 27 ms⁻¹ when t = 2.

Mark scheme

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

1. [M1*] Attempt to integrate a. The power of t must increase by 1 with a change of coefficient in the t² term. Do not penalise missing c. Use of v = at scores M0. (e.g., [v=]36t-3t² [+c])
2. [A1] Condone an integral sign in front of correct answer. (e.g., [v=]36t-6t²/2 [+c])
3. [DM1] Use t=2 and v = 27 to find c. Must get to c = and set 3 term quadratic equal to zero. (e.g., 27=36×2-3×2²+c ⇒c=-33)
4. [A1] Solve 0=36t-3t² -33 to get t =1 and t=11; t=1, t=11 oe.
5. [M1*] Attempt to integrate an expression of the form at +bt² [+c] with non-zero a and b. The power of t must increase by 1 with a change of coefficient in the same term. Use of s = vt scores M0. (e.g., [s=]18t² -t³ -33t [+c'])
6. [DM1] Attempt to evaluate their 18t²-t³-33t for t=0 to t =1 or t=1 to t=11 or t=11 to t=12. Attempt using their limits (at least one strictly between 0 and 12) correctly. (e.g., 0 to 1:-16-0 = -16, 1 to 11: 484 – (-16) = 500, 11 to 12: 468 – 484 = -16)
7. [DM1] For all three. Allow 11 to 12 implied by symmetry instead of found separately.
8. [A1] Distance = [16+500+16]=532m