M2 June 2011 Q6
6. A particle \(P\) moves on the \(x\)-axis. The acceleration of \(P\) at time \(t\) seconds is \((t - 4)\) m s\(^{-2}\) in the positive \(x\)-direction. The velocity of \(P\) at time \(t\) seconds is \(v\) m s\(^{-1}\). When \(t = 0\), \(v = 6\).
Find
(a) \(v\) in terms of \(t\), (4)
(b) the values of \(t\) when \(P\) is instantaneously at rest, (3)
(c) the distance between the two points at which \(P\) is instantaneously at rest. (4)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} = t - 4\) | |
| \(v = \dfrac{1}{2}t^2 - 4t\ (+c)\) | M1 A1 |
| \(t = 0\ \ v = 6\ \ \Rightarrow c = 6\) | M1 |
| \(\therefore v = \dfrac{1}{2}t^2 - 4t + 6\) | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(v = 0\quad 0 = t^2 - 8t + 12\) | M1 |
| \((t - 6)(t - 2) = 0\) | DM1 |
| \(t = 6\quad t = 2\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(x = \dfrac{t^3}{6} - 2t^2 + 6t + k\) | M1 A1 ft |
| \(x_6 - x_2 = \dfrac{6^3}{6} - 2 \times 6^2 + 6^2 + k - \left(\dfrac{2^3}{6} - 2 \times 2^2 + 6 \times 2 + k\right)\) | DM1 |
| \(= -5\dfrac{1}{3}\) | |
| \(\therefore\) Distance is \(5\dfrac{1}{3}\) m | A1 |
| (4) | |
| (11 marks) |