M4 June 2010 Q3
3. At 12 noon, ship \(A\) is 8 km due west of ship \(B\). Ship \(A\) is moving due north at a constant speed of 10 km h\(^{-1}\). Ship \(B\) is moving at a constant speed of 6 km h\(^{-1}\) on a bearing so that it passes as close to \(A\) as possible.
(a) Find the bearing on which ship \(B\) moves. (4)
(b) Find the shortest distance between the two ships. (3)
(c) Find the time when the two ships are closest. (3)
| Scheme | Marks |
|---|---|
 | M1 |
| \(\cos\theta = \dfrac{6}{10} \Rightarrow \theta = 53.1^\circ\) | M1 A1 |
| Bearing is \(307^\circ\) | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(d = 8\sin\theta\ (= 8 \times 0.8)\) | M1 A1 |
| \(= 6.4\) km | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(T = \dfrac{8\cos\theta}{\sqrt{10^2 - 6^2}}\) | M1 A1 |
| \(= 0.6\) hrs i.e. the time is 12:36 pm | A1 |
| (3) | |
| (10 marks) |