M1 January 2006 Q6
6. [In this question the horizontal unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are due east and due north respectively.]
A model boat \(A\) moves on a lake with constant velocity \((-\mathbf{i} + 6\mathbf{j})\) m s\(^{-1}\). At time \(t = 0\), \(A\) is at the point with position vector \((2\mathbf{i} - 10\mathbf{j})\) m. Find
(a) the speed of \(A\), (2)
(b) the direction in which \(A\) is moving, giving your answer as a bearing. (3)
At time \(t = 0\), a second boat \(B\) is at the point with position vector \((-26\mathbf{i} + 4\mathbf{j})\) m.
Given that the velocity of \(B\) is \((3\mathbf{i} + 4\mathbf{j})\) m s\(^{-1}\),
(c) show that \(A\) and \(B\) will collide at a point \(P\) and find the position vector of \(P\). (5)
Given instead that \(B\) has speed 8 m s\(^{-1}\) and moves in the direction of the vector \((3\mathbf{i} + 4\mathbf{j})\),
(d) find the distance of \(B\) from \(P\) when \(t = 7\) s. (6)
| Scheme | Marks |
|---|---|
| Speed of \(A = \sqrt{(1^2 + 6^2)} \approx 6.08\) m s\(^{-1}\) | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\tan\theta = 1/6 \;\Rightarrow\; \theta \approx 9.46^\circ\) | M1 A1 |
| Bearing \(\approx 351\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| P.v. of \(A\) at time \(t = (2 - t)\mathbf{i} + (-10 + 6t)\mathbf{j}\) p.v. of \(B\) at time \(t = (-26 + 3t)\mathbf{i} + (4 + 4t)\mathbf{j}\) | B1 (either) |
| (E.g.) \(\mathbf{i}\) components equal \(\Rightarrow 2 - t = -26 + 3t \;\Rightarrow\; t = 7\) | M1 A1 |
| \(\mathbf{j}\) components at \(t = 7\): \(A\): \(-10 + 6t = 32\) \(B\): \(4 + 4t = 32\) | M1 |
| Same, so collide at \(t = 7\) s at point with p.v. \((-5\mathbf{i} + 32\mathbf{j})\) m | A1 cso |
| (5) |
| Scheme | Marks |
|---|---|
| New velocity of \(B = \dfrac{8}{5}(3\mathbf{i} + 4\mathbf{j})\) m s\(^{-1}\) | B1 |
| P.v. of \(B\) at 7 s \(= -26\mathbf{i} + 4\mathbf{j} + 1.6(3\mathbf{i} + 4\mathbf{j}) \times 7 = 7.6\mathbf{i} + 48.8\mathbf{j}\) | M1 A1 |
| \(\overline{PB} = \mathbf{b} - \mathbf{p} = 12.6\mathbf{i} + 16.8\mathbf{j}\) (in numbers) | M1 |
| Distance \(= \sqrt{(12.6^2 + 16.8^2)} = 21\) m | M1 A1 |
| (6) | |
| (16 marks) |