M4 January 2005 Q3
3. Two ships \(A\) and \(B\) are sailing in the same direction at constant speeds of 12 km h\(^{-1}\) and 16 km h\(^{-1}\) respectively. They are sailing along parallel lines which are 4 km apart. When the distance between the ships is 4 km, \(B\) turns through 30\(^\circ\) towards \(A\).
Find the shortest distance between the ships in the subsequent motion. (7)
| Scheme |
|---|
| Treat \(B\) when \(t = 0\) as the origin. |
| \(\mathbf{r}_A = 12t\mathbf{i} + 4\mathbf{j}\). \(\quad\mathbf{r}_B = 16t\left(\dfrac{\sqrt{3}}{2}\mathbf{i} + \dfrac{1}{2}\mathbf{j}\right)\) |
| \(\mathbf{BA} = \mathbf{r}_A - \mathbf{r}_B = \mathbf{i}\left(12t - 8t\sqrt{3}\right) + \mathbf{j}(4 - 8t)\) |
| Length of \(\mathbf{AB} = \sqrt{\left(\left(12t - 8t\sqrt{3}\right)^2 + (4 - 8t)^2\right)} = \sqrt{\left(\left(144 - 192\sqrt{3} + 192\right)t^2 + 16 - 64t + 64t^2\right)}\) |
| Minimum when derivative of terms inside square root = 0: |
| \(2t\left(144 - 192\sqrt{3} + 192\right) - 64 + 128t = 0,\ \ t \approx 0.47\). (Minimum because this is a +ve quadratic.) |
| Substitute back into length of \(\mathbf{AB}\): \(\ \left|\mathbf{AB}\right| \approx 0.90\) km. |
Notes
The published mark scheme for this paper is a set of worked answers: no mark allocation is printed.
(Corrected from the printed mark scheme: the position vector of \(B\) is printed as \(\mathbf{r}_A = 16t\left(\dfrac{\sqrt{3}}{2}\mathbf{i} + \dfrac{1}{2}\mathbf{j}\right)\).)