M5 June 2007 Q2
2. At time \(t\) seconds, the position vector of a particle \(P\) is \(\mathbf{r}\) metres, where \(\mathbf{r}\) satisfies the differential equation
\[\frac{\mathrm{d}^2\mathbf{r}}{\mathrm{d}t^2} + 3\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} = \mathbf{0}.\]When \(t = 0\), the velocity of \(P\) is \((8\mathbf{i} - 12\mathbf{j})\) m s\(^{-1}\).
Find the velocity of \(P\) when \(t = \tfrac{2}{3}\ln 2\).
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} + 3\mathbf{v} = \mathbf{0}\) | B1 |
| IF \(= \mathrm{e}^{3t} \Rightarrow \dfrac{\mathrm{d}(\mathbf{v}\mathrm{e}^{3t})}{\mathrm{d}t} = \mathbf{0}\) | M1 |
| \(\Rightarrow\ \mathbf{v}\mathrm{e}^{3t} = \mathbf{A}\) | A1 |
| \(t = 0,\ \mathbf{v} = 8\mathbf{i} - 12\mathbf{j} \Rightarrow \mathbf{v} = (8\mathbf{i} - 12\mathbf{j})\mathrm{e}^{-3t}\) | M1 A1 |
| \(t = \tfrac{2}{3}\ln 2 \Rightarrow \mathbf{v} = (8\mathbf{i} - 12\mathbf{j})\mathrm{e}^{-2\ln 2} = (2\mathbf{i} - 3\mathbf{j})\) m s\(^{-1}\) | DM1 A1 |
| (7) |