M4 June 2013 Q5
5. A coastguard ship \(C\) is due south of a ship \(S\). Ship \(S\) is moving at a constant speed of 12 km h\(^{-1}\) on a bearing of 140\(^\circ\). Ship \(C\) moves in a straight line with constant speed \(V\) km h\(^{-1}\) in order to intercept \(S\).
(a) Find, giving your answer to 3 significant figures, the minimum possible value for \(V\). (3)
It is now given that \(V = 14\)
(b) Find the bearing of the course that \(C\) takes to intercept \(S\). (5)
| Scheme | Marks |
|---|---|
| Minimum \(V = 12\cos 50^\circ\) | M1 A1 |
| \(\approx 7.71\) | A1 |
Notes
M1 Use of triangle with right angle between \(v_C\) and \({}_Cv_S\). Condone sin/cos confusion.
A1 Correct unsimplified trig expression
A1 7.71 only
| Scheme | Marks |
|---|---|
| Vector triangle for relative velocities when \(V = 14\) | M1 |
| Select the vector triangle with the relative velocity due N. | A1 |
| \(\dfrac{\sin\theta}{12} = \dfrac{\sin 40}{14}\) | DM1 A1 |
| Bearing 033\(^\circ\) | A1 |
| (8 marks) |
Notes
M1 could have relative velocity due S. Could show both possibilities.
DM1 Use of sine rule or equivalent to find \(\theta\)
A1 Final answer. Accept 33.4\(^\circ\)