M1 June 2009 Q8
8. [In this question \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors due east and due north respectively.]
A hiker \(H\) is walking with constant velocity \((1.2\mathbf{i} - 0.9\mathbf{j})\) m s\(^{-1}\).
(a) Find the speed of \(H\). (2)
A horizontal field \(OABC\) is rectangular with \(OA\) due east and \(OC\) due north, as shown in Figure 3. At twelve noon hiker \(H\) is at the point \(Y\) with position vector \(100\mathbf{j}\) m, relative to the fixed origin \(O\).
(b) Write down the position vector of \(H\) at time \(t\) seconds after noon. (2)
At noon, another hiker \(K\) is at the point with position vector \((9\mathbf{i} + 46\mathbf{j})\) m. Hiker \(K\) is moving with constant velocity \((0.75\mathbf{i} + 1.8\mathbf{j})\) m s\(^{-1}\).
(c) Show that, at time \(t\) seconds after noon,\[\overrightarrow{HK} = \left[(9 - 0.45t)\mathbf{i} + (2.7t - 54)\mathbf{j}\right]\text{ metres.}\](4)
Hence,
(d) show that the two hikers meet and find the position vector of the point where they meet. (5)
| Scheme | Marks |
|---|---|
| \(|\mathbf{v}| = \sqrt{1.2^2 + (-0.9)^2} = 1.5\) m s\(^{-1}\) | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \((\mathbf{r}_H =)\,100\mathbf{j} + t(1.2\mathbf{i} - 0.9\mathbf{j})\) m | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \((\mathbf{r}_K =)\,9\mathbf{i} + 46\mathbf{j} + t(0.75\mathbf{i} + 1.8\mathbf{j})\) m | M1 A1 |
| \(\overrightarrow{HK} = \mathbf{r}_K - \mathbf{r}_H = (9 - 0.45t)\mathbf{i} + (2.7t - 54)\mathbf{j}\) m Printed Answer | M1 A1 |
| (4) |
| Scheme | Marks |
|---|---|
| Meet when \(\overrightarrow{HK} = \mathbf{0}\) | |
| \((9 - 0.45t) = 0\) and \((2.7t - 54) = 0\) | M1 A1 |
| \(t = 20\) from both equations | A1 |
| \(\mathbf{r}_K = \mathbf{r}_H = (24\mathbf{i} + 82\mathbf{j})\) m | DM1 A1 cso |
| (5) | |
| (13 marks) |