M4 June 2013 (R) Q4
4. At 10 a.m. two walkers \(A\) and \(B\) are 4 km apart with \(A\) due north of \(B\). Walker \(A\) is moving due east at a constant speed of 6 km h\(^{-1}\). Walker \(B\) is moving with constant speed 5 km h\(^{-1}\) and walks in the straight line which allows him to pass as close as possible to \(A\).
Find
(a) the direction of motion of \(B\), giving your answer as a bearing, (4)
(b) the least distance between \(A\) and \(B\), (2)
(c) the time when the distance between \(A\) and \(B\) is least. (4)
| Scheme | Marks |
|---|---|
 | B1 |
| \(\sin\theta = \dfrac{5}{6}\) | M1 |
| \(\theta = 56.44\ldots\) | A1 |
| Bearing \(= 056^\circ\) | A1 |
| (4) |
Notes
B1 Right angled triangle with the right angle opposite the 6 seen in diagram or implied in working
M1 Correct trig.
A1 Allow 56.4\(^\circ\)
| Scheme | Marks |
|---|---|
| Least distance \(= 4\cos\theta = \dfrac{\left(4\sqrt{11}\right)}{6}\) or 2.211 km oe | M1 A1 |
| (2) |
Notes
M1 Correct for their angle
A1 2.2 or better
| Scheme | Marks |
|---|---|
| \({}_Bv_A = \sqrt{6^2 - 5^2} = \sqrt{11}\) | B1 |
| \(t = \dfrac{4\sin\theta}{\sqrt{11}} \quad (= 1.0050\ldots)\) | M1 A1ft |
| time = 11 am | B1 |
| (4) | |
| (10 marks) |
Notes
B1 3.32
M1 Condone consistent trig confusion
A1ft Ft on their \(\sqrt{11}\)