M4 June 2015 Q2
2. When a woman walks due North at a constant speed of 4 km h\(^{-1}\), the wind appears to be blowing from due East. When she runs due South at a constant speed of 8 km h\(^{-1}\), the speed of the wind appears to be 20 km h\(^{-1}\).
Assuming that the velocity of the wind relative to the earth is constant, find
(i) the speed of the wind,
(ii) the direction from which the wind is blowing. (6)
| Scheme | Marks |
|---|---|
 | B1 M1 A1 |
| Correct method to obtain one of \(v, w, \theta\) | M1 |
| speed is 16.5(km h\(^{-1}\)) | A1 |
| Direction S\(76^\circ\)E or equivalent | A1 |
| (6) | |
| (6 marks) |
Notes
B1 Either triangle of velocities
M1 Two triangles combined using their common velocity
A1 Correct diagram seen or implied
M1 \((v = 16,\ w = 16.5,\ \theta = 76^\circ)\) Make it dependent?
A1 \(4\sqrt{17}\)
A1 104\(^\circ\)
Alt
| Velocity of wind \(= w\) | |
| \(w = -v\mathbf{i} + 4\mathbf{j}\) | B1 |
| \(w = a\mathbf{i} + b\mathbf{j} - 8\mathbf{j} \qquad a^2 + b^2 = 400\) | M1 |
| coeff \(\mathbf{j}\): \(\ 4 = b - 8 \qquad b = 12\) \(\phantom{coeff}\ \mathbf{i}\): \(\ -v = a\) | A1 |
| \(a^2 + 144 = 400 \Rightarrow a = -16 \quad (v > 0)\) | M1 |
| \(|w| = \sqrt{4^2 + 16^2} = 4\sqrt{17}\) | A1 |
| Bearing 104\(^\circ\) | A1 |
B1 one correct equation
M1 2nd equation and compare coefficients
A1 2 correct eqns