M1 January 2010 Q7
7. [In this question, \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors due east and due north respectively and position vectors are given with respect to a fixed origin.]
A ship \(S\) is moving along a straight line with constant velocity. At time \(t\) hours the position vector of \(S\) is \(\mathbf{s}\) km. When \(t = 0\), \(\mathbf{s} = 9\mathbf{i} - 6\mathbf{j}\). When \(t = 4\), \(\mathbf{s} = 21\mathbf{i} + 10\mathbf{j}\). Find
(a) the speed of \(S\), (4)
(b) the direction in which \(S\) is moving, giving your answer as a bearing. (2)
(c) Show that \(\mathbf{s} = (3t + 9)\mathbf{i} + (4t - 6)\mathbf{j}\). (2)
A lighthouse \(L\) is located at the point with position vector \((18\mathbf{i} + 6\mathbf{j})\) km. When \(t = T\), the ship \(S\) is 10 km from \(L\).
(d) Find the possible values of \(T\). (6)
| Scheme | Marks |
|---|---|
| \(\mathbf{v} = \dfrac{21\mathbf{i} + 10\mathbf{j} - (9\mathbf{i} - 6\mathbf{j})}{4} = 3\mathbf{i} + 4\mathbf{j}\) | M1 A1 |
| speed is \(\sqrt{\left(3^2 + 4^2\right)} = 5\ \ (\text{km h}^{-1})\) | M1 A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\tan\theta = \dfrac{3}{4}\ \ (\Rightarrow \theta \approx 36.9^\circ)\) | M1 |
| bearing is 37, 36.9, 36.87, … | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathbf{s} = 9\mathbf{i} - 6\mathbf{j} + t(3\mathbf{i} + 4\mathbf{j})\) | M1 |
| \(= (3t + 9)\mathbf{i} + (4t - 6)\mathbf{j}\) * cso | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| Position vector of \(S\) relative to \(L\) is | |
| \((3T + 9)\mathbf{i} + (4T - 6)\mathbf{j} - (18\mathbf{i} + 6\mathbf{j}) = (3T - 9)\mathbf{i} + (4T - 12)\mathbf{j}\) | M1 A1 |
| \((3T - 9)^2 + (4T - 12)^2 = 100\) | M1 |
| \(25T^2 - 150T + 125 = 0\) or equivalent | DM1 A1 |
| \(\left(T^2 - 6T + 5 = 0\right)\) | |
| \(T = 1, 5\) | A1 |
| (6) | |
| (14 marks) |