M4 June 2012 Q4
4. A rescue boat, whose maximum speed is 20 km h\(^{-1}\), receives a signal which indicates that a yacht is in distress near a fixed point \(P\). The rescue boat is 15 km south-west of \(P\). There is a constant current of 5 km h\(^{-1}\) flowing uniformly from west to east. The rescue boat sets the course needed to get to \(P\) as quickly as possible. Find
When the rescue boat arrives at \(P\), the yacht is just visible 4 km due north of \(P\) and is drifting with the current. Find
| Scheme | Marks |
|---|---|
 | |
| \(\dfrac{\sin\theta}{5} = \dfrac{\sin 45}{20}\) | M1 A1 |
| \(\theta = 10.182\ldots\) | |
| Bearing is \(45^\circ - \theta = 34.8 = 35^\circ\) (nearest degree) | M1 A1 |
| (4) |
Notes
M1 Use a vector triangle to find \(\theta\). Condone the 5 ms\(^{-1}\) in the wrong direction.
A1 Correct equation for \(\theta\)
M1 Use their angle correctly in their triangle to find the bearing.
A1 Accept alternative forms e.g. N 35 E
OR
| SW \(\rightarrow (20\sin\theta)T = (5 + 20\cos\theta)T\) | M1 |
| \(3t^2 + 8t - 5 = 0,\ \ t = \dfrac{-8 + \sqrt{124}}{6} = 0.5225\ldots\) | A1 |
| \(\theta = 55.18\ldots\) Bearing is \(90 - \theta = 34.8^\circ\) | M1 A1 |
M1 \(45^\circ\) rt angle triangle
A1 \(t\) substitution leading to correct equation in \(t\), use of \(R\cos(\theta + \alpha)\) o.e.
| Scheme | Marks |
|---|---|
| \(v^2 = 5^2 + 20^2 - 2 \times 5 \times 20\cos 124.818\ldots\) OR \(v = \dfrac{20}{\sin 45} \times \sin 124.8\) OR \(v = 5\cos 45 + 20\cos\theta\) | M1 |
| \(v = 23.22\) | A1 |
| \(t = \dfrac{15}{23.22} = 0.646\) h \(= 39\) min (nearest min) | M1 A1 |
| (4) |
Notes
M1 Complete method to find \(v\)
A1 Or better \(\left(\dfrac{5\sqrt{2} + 5\sqrt{62}}{2}\right)\)
M1 \(\dfrac{15}{\text{their } v}\)
A1 The Q specifies “nearest minute”
| Scheme | Marks |
|---|---|
| Due N, (since current affects both equally) | B1 |
| (1) |
Notes
B1 cao
| Scheme | Marks |
|---|---|
| \(t = \dfrac{4}{20} = 0.2\) h \(= 12\) min | B1 |
| (1) | |
| (10 marks) |
Notes
B1 cso