M4 January 2005 Q2
2. [In this question \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors due east and due north respectively.]
A man cycling at a constant speed \(u\) on horizontal ground finds that, when his velocity is \(u\mathbf{j}\) m s\(^{-1}\), the velocity of the wind appears to be \(v(3\mathbf{i} - 4\mathbf{j})\) m s\(^{-1}\), where \(v\) is a constant. When the velocity of the man is \(\dfrac{u}{5}(-3\mathbf{i} + 4\mathbf{j})\) m s\(^{-1}\), he finds that the velocity of the wind appears to be \(w\mathbf{i}\) m s\(^{-1}\), where \(w\) is a constant.
| Scheme |
|---|
| Velocity of wind relative to man \(= \mathbf{V}_{WM} = \mathbf{V}_W - \mathbf{V}_M\). \(\quad\therefore v(3\mathbf{i} - 4\mathbf{j}) = \mathbf{V}_W - u\mathbf{j}\) |
| Similarly \(w\mathbf{i} = \mathbf{V}_W - \tfrac{1}{5}u(-3\mathbf{i} + 4\mathbf{j})\). |
| Equate the two expressions for \(\mathbf{V}_W\) that these produce: \(v(3\mathbf{i} - 4\mathbf{j}) + u\mathbf{j} = w\mathbf{i} + \tfrac{1}{5}u(-3\mathbf{i} + 4\mathbf{j})\) |
| Equate coefficients: \(\quad\mathbf{i}\colon\ 3v = w - \tfrac{3}{5}u\) |
| \(\mathbf{j}\colon\ -4v + u = \tfrac{4}{5}u \qquad \therefore v = \tfrac{1}{20}u\) |
Notes
The published mark scheme for this paper is a set of worked answers: no mark allocation is printed.
(Corrected from the printed mark scheme: the printed answers say “the two expressions for \(\mathbf{V}_M\)” and give the \(\mathbf{i}\) equation as \(-3v = w - \tfrac{3}{5}u\).)
The printed answers do not state \(w\); from the \(\mathbf{i}\) equation, \(w = 3v + \tfrac{3}{5}u = \tfrac{3}{20}u + \tfrac{12}{20}u = \tfrac{3}{4}u\).
| Scheme |
|---|
| \(\mathbf{V}_W = \tfrac{1}{20}u(3\mathbf{i} - 4\mathbf{j}) + u\mathbf{j} = \tfrac{1}{20}u(3\mathbf{i} + 16\mathbf{j})\) |