M4 June 2014 Q4
4. At noon two ships \(A\) and \(B\) are 20 km apart with \(A\) on a bearing of 230\(^\circ\) from \(B\). Ship \(B\) is moving at 6 km h\(^{-1}\) on a bearing of 015\(^\circ\). The maximum speed of \(A\) is 12 km h\(^{-1}\). Ship \(A\) sets a course to intercept \(B\) as soon as possible.
(a) Find the course set by \(A\), giving your answer as a bearing to the nearest degree. (4)
(b) Find the time at which \(A\) intercepts \(B\). (4)
| Scheme | Marks |
|---|---|
| Relative velocity triangle | M1 |
| \(\dfrac{\sin 145}{12} = \dfrac{\sin\theta}{6},\ \ \theta = 16.7^\circ\) | M1 |
| Bearing \(= 15 + (180 - 145 - 16.7) = 33.3^\circ\) | M1 |
| Bearing 033\(^\circ\) | A1 |
| (4) |
Notes
M1 Seen or implied
M1 Use of trig to find a relevant angle
M1 To find the required angle
A1 They were asked for an answer "to the nearest degree". Accept N 33\(^\circ\) E
| Scheme | Marks |
|---|---|
| \(\dfrac{{}_Av_B}{\sin 18.3} = \dfrac{12}{\sin 145}\) | M1 |
| \({}_Av_B = 6.58\) (km h\(^{-1}\)) | A1 |
| Time taken \(= \dfrac{20}{6.58}\) (hrs) | M1 |
| Time is 3:02 pm (1502) | A1 |
| (4) | |
| (8 marks) |
Notes
M1 Correct method to find the relative velocity
M1 For their 6.58