M4 June 2018 Q3
3. When a man walks due West at a constant speed of 4 km h\(^{-1}\), the wind appears to be blowing from due South. When he runs due North at a constant speed of 8 km h\(^{-1}\), the speed of the wind appears to be 5 km h\(^{-1}\).
The velocity of the wind relative to the Earth is constant with magnitude \(w\) km h\(^{-1}\).
Find the two possible values of \(w\). (7)
| Scheme | Marks |
|---|---|
| \(\mathbf{v}_w = {}_{w}\mathbf{v}_m + \mathbf{v}_m\) walking: \(\mathbf{v}_w = a\mathbf{j} - 4\mathbf{i}\) | B1 |
| Running: \(\mathbf{v}_w = b\mathbf{i} + (c + 8)\mathbf{j}\quad \left(b^2 + c^2 = 25\right)\) | B1 |
| Compare components and use \(b^2 + c^2 = 25\): | M1 |
| \(b = -4\) | A1 |
| \(a = c + 8,\ c^2 = 25 - 16 = 9,\ c = \pm 3\) | A1 |
| Correct method to obtain a value of \(w\): \(w = \sqrt{4^2 + 5^2} = \sqrt{41}\ (= 6.40)\) | M1 |
| Second value correct : \(w = \sqrt{4^2 + 11^2} = \sqrt{137}\ (= 11.7)\) | A1 |
| (7) | |
| (7 marks) |
Alternative
| Triangle of velocities for walking | B1 |
| Either form of triangle of velocities for running using their \(v_w\) | B1 |
| Two triangles combined using their common velocity | M1 |
| Either correct diagram seen or implied | A1 |
| Both possibilities shown | A1 |
| Correct method to obtain a value of \(w\): \(w = \sqrt{4^2 + 5^2} = \sqrt{41}\ (= 6.40)\) | M1 |
| Second value correct : \(w = \sqrt{4^2 + 11^2} = \sqrt{137}\ (= 11.7)\) | A1 |
| (7) |