M5 June 2010 Q1
1. At time \(t = 0\), the position vector of a particle \(P\) is \(-3\mathbf{j}\) m. At time \(t\) seconds, the position vector of \(P\) is \(\mathbf{r}\) metres and the velocity of \(P\) is \(\mathbf{v}\) m s\(^{-1}\). Given that
\[\mathbf{v} - 2\mathbf{r} = 4\mathrm{e}^{t}\mathbf{j},\]find the time when \(P\) passes through the origin. (7)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} - 2\mathbf{r} = 4\mathrm{e}^{t}\mathbf{j}\) IF \(= \mathrm{e}^{-2t}\) | |
| \(\mathrm{e}^{-2t}\left(\dfrac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} - 2\mathbf{r}\right) = \mathrm{e}^{-2t}.4\mathrm{e}^{t}\mathbf{j}\) | M1 |
| \(\dfrac{\mathrm{d}(\mathbf{r}\mathrm{e}^{-2t})}{\mathrm{d}t} = 4\mathrm{e}^{-t}\mathbf{j}\) \(\mathbf{r}\mathrm{e}^{-2t} = \displaystyle\int 4\mathrm{e}^{-t}\mathbf{j}\,\mathrm{d}t\) | DM1 |
| \(= -4\mathrm{e}^{-t}\mathbf{j}\ (+\ \mathbf{C})\) | A1 |
| \(t = 0,\ \mathbf{r} = -3\mathbf{j} \Rightarrow \mathbf{C} = \mathbf{j}\) | DM1 |
| \(\mathrm{e}^{-2t}\mathbf{r} = (1 - 4\mathrm{e}^{-t})\mathbf{j}\) or \(\mathbf{r} = (\mathrm{e}^{2t} - 4\mathrm{e}^{t})\mathbf{j}\) | A1 |
| \((1 - 4\mathrm{e}^{-t}) = 0\) or \((\mathrm{e}^{2t} - 4\mathrm{e}^{t}) = 0\) | DM1 |
| \(t = \ln 4\), 1.4 or better | A1 |
| (7 marks) |