M4 January 2006 Q3
3. Two ships \(P\) and \(Q\) are moving with constant velocity. At 3 p.m., \(P\) is 20 km due north of \(Q\) and is moving at 16 km h\(^{-1}\) due west. To an observer on ship \(P\), ship \(Q\) appears to be moving on a bearing of 030\(^\circ\) at 10 km h\(^{-1}\). Find
(a)
(i) the speed of \(Q\),
(ii) the direction in which \(Q\) is moving, giving your answer as a bearing to the nearest degree, (6)
(b) the shortest distance between the ships, (3)
(c) the time at which the two ships are closest together. (3)
| Scheme | Marks |
|---|---|
| (i) \(\ \mathbf{v}_Q = {}_Q\mathbf{v}_P + \mathbf{v}_P\) | |
| \(|\mathbf{v}_Q|^2 = (10\cos 30)^2 + (16 - 10\sin 30)^2\) | M1 A1 |
| \(= 75 + 121\) | |
| \(\Rightarrow |\mathbf{v}_Q| = 14\,ms^{-1}\) | A1 |
| (ii) \(\ \tan\theta = \dfrac{16 - 10\sin 30}{10\cos 30}\qquad\) (o.e.) | M1 |
| \(\theta \approx 51.8^\circ,\ \Rightarrow\) bearing \(308^\circ\) (nearest degree) | A1, A1 |
| (6) |
Notes
(Corrected from the printed mark scheme: the numerator is printed as \(16 - \sin 30\).)
Alternatives
(a) Use of cosine rule in velocity vector triangle.
| Scheme | Marks |
|---|---|
| At nearest approach: \(\ PN = 20\sin 30\) | M1 A1 |
| \(= 10\ km\) | A1 |
| (3) |
Alternatives
(b) & (c) Use of scalar product of relative velocity and relative position or differentiating magnitude of relative position vector squared to find the minimum distance and time at which it occurs.
| Scheme | Marks |
|---|---|
| \(Time = \dfrac{20\cos 30}{10} \approx 1.732\ hrs\) | M1 A1 |
| \(\Rightarrow Time \approx 4.44\ pm\qquad\) (AWRT) | A1 |
| (3) | |
| (12 marks) |