A particle is moving in a straight line such that its velocity, v ms⁻¹, t seconds after passing a fixed point O is v = e^(2t) - 6e^(-2t) - 1.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2015). The mark or mark type is shown in brackets.

1. [M1] (i) s = 1/2 e^(2t) + 3e^(-2t) - t + (c) Integrate : coefficient of 1/2 or 3 seen with no change in powers of e. Ignore –t
2. [A1] (i) t = 0, s = 0 → c = -3.5
3. [A1] (i) s = 1/2 e^(2t) + 3e^(-2t) - t - 3.5 All correct and simplified
4. [M1] (ii) v = 0 → u² - u - 6 = 0 oe Obtain three term quadratic in u or e^(2t) Condone sign errors.
5. [DM1] (ii) (u - 3)(u + 2) = 0 Solve three term quadratic
6. [A1] (ii) → u = 3 → t = 1/2 ln 3 or 0.549 Accept 0.55 No second answer
7. [B1] (iii) t = 1/2 ln 3 → a = 2e^(2t) + 12e^(-2t) Correct differentiation
8. [B1] (iii) = 6 + 4 = 10 Allow awrt 10.0 or 9.99. No second answer.