M4 June 2011 Q3
3. [In this question the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are due east and due north respectively.]
A coastguard patrol boat \(C\) is moving with constant velocity \((8\mathbf{i} + u\mathbf{j})\) km h\(^{-1}\). Another ship \(S\) is moving with constant velocity \((12\mathbf{i} + 16\mathbf{j})\) km h\(^{-1}\).
(a) Find, in terms of \(u\), the velocity of \(C\) relative to \(S\). (2)
At noon, \(S\) is 10 km due west of \(C\).
If \(C\) is to intercept \(S\),
(b)
(i) find the value of \(u\).
(ii) Using this value of \(u\), find the time at which \(C\) would intercept \(S\).
(4)If instead, at noon, \(C\) is moving with velocity \((8\mathbf{i} + 8\mathbf{j})\) km h\(^{-1}\) and continues at this constant velocity,
(c) find the distance of closest approach of \(C\) to \(S\). (5)
| Scheme | Marks |
|---|---|
| Velocity of C relative to S \(= (8\mathbf{i} + u\mathbf{j}) - (12\mathbf{i} + 16\mathbf{j})\) | M1 |
| \(= (-4\mathbf{i} + (u - 16)\mathbf{j})\) (km h\(^{-1}\)) | A1 |
| (2) |
Notes
(Corrected from the printed mark scheme: the units are printed as m s\(^{-1}\).)
| Scheme | Marks |
|---|---|
| (i) C intercepts S \(\Rightarrow\) relative velocity is parallel to \(\mathbf{i}\). | |
| \(\Rightarrow u - 16 = 0,\ u = 16\) | M1 A1 |
| (2) | |
| (ii) 10 km at 4 km h\(^{-1}\) takes 2.5 hours, so 2.30pm | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(u = 8\), relative velocity \(= -4\mathbf{i} - 8\mathbf{j}\). | B1 |
 | |
| Correct distance identified | B1 |
| Using velocity: \(\ \tan\theta = \dfrac{8}{4} = 2 \Rightarrow \sin\theta = \dfrac{2}{\sqrt{5}}\) | |
| Using distance: \(\ \sin\theta = \dfrac{d}{10} = \dfrac{2}{\sqrt{5}}\), | M1 A1 |
| \(d = \dfrac{20}{\sqrt{5}} = 4\sqrt{5} = 8.9\) (km) | A1 |
| (5) | |
| (11 marks) |