M4 June 2017 Q1
1. [In this question the horizontal unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are due east and due north respectively.]
A ship \(A\) has constant velocity \((4\mathbf{i} + 2\mathbf{j})\) km h\(^{-1}\) and a ship \(B\) has constant velocity \((-\mathbf{i} + 3\mathbf{j})\) km h\(^{-1}\). At noon, the position vectors of the ships \(A\) and \(B\) with respect to a fixed origin \(O\) are \((-2\mathbf{i} + \mathbf{j})\) km and \((5\mathbf{i} - 2\mathbf{j})\) km respectively.
Find
| Scheme | Marks |
|---|---|
| Position vectors after \(t\) hours: \(\begin{pmatrix}-2+4t\\1+2t\end{pmatrix}\) and \(\begin{pmatrix}5-t\\-2+3t\end{pmatrix}\) | M1 |
| \(\pm\begin{pmatrix}7-5t\\-3+t\end{pmatrix}\) | A1 |
| \(d^2 = (7 - 5t)^2 + (-3 + t)^2\) \(= 26t^2 - 76t + 58\) | M1 |
| Differentiate: \(52t - 76 = 0\) | DM1 |
| \(t = \dfrac{76}{52} = 1.46\) hrs, 13.28 | A1 |
| (5) |
Notes
M1 Use position vectors to find position of one ship relative to the other
M1 Correct method for magnitude
DM1 or complete the square
1a alt
| Position vectors after \(t\) hours: \(\begin{pmatrix}-2+4t\\1+2t\end{pmatrix}\) and \(\begin{pmatrix}5-t\\-2+3t\end{pmatrix}\) | M1 |
| \(\pm\begin{pmatrix}7-5t\\-3+t\end{pmatrix}\) | A1 |
| At closest point: \(\begin{pmatrix}5\\-1\end{pmatrix}\cdot\begin{pmatrix}7-5t\\-3+t\end{pmatrix} = 0,\) \(5(7 - 5t) - (-3 + t) = 0\) | M1 |
| \(35 - 25t + 3 - t = 0\) | DM1 |
| \(t = \dfrac{38}{26} = 1.46\) hrs, 13.28 | A1 |
| (5) |
M1 Use position vectors to find position of one ship relative to the other
M1 Scalar product of relative velocity and relative position
DM1 Scalar product = 0
1a alt
| Angle between initial positions of \(A\) and \(B\) and relative velocity \(= \theta = \tan^{-1}\left(\dfrac{3}{7}\right) - \tan^{-1}\left(\dfrac{1}{5}\right)\) | |
| \(= 11.89^\circ\) | M1A1 |
| \(d = \sqrt{58}\cos\theta\) | M1 |
| Time taken \(= \dfrac{d}{\sqrt{26}}\) | DM1 |
| \(t = 1.46\) hrs, 13.28 | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| Distance \(\leqslant 2: d^2 = 26t^2 - 76t + 58 \leqslant 4,\) \(26t^2 - 76t + 54 \leqslant 0\) | M1 |
| Time interval: \(2\times\dfrac{\sqrt{76^2 - 4\times 26\times 54}}{52}\ \left(= 2\dfrac{\sqrt{160}}{52}\right)\) | M1 |
| \(= 0.487\) hrs (29 mins) | A1 |
| (3) | |
| (8 marks) |
Notes
M1 (condone equality)
M1 Difference between roots
A1 0.49 or better
1b alt
| \(\dfrac{\sin\alpha}{\sqrt{58}} = \dfrac{\sin 11.89}{2} \Rightarrow \alpha = 51.7^\circ, 128.3^\circ\) \(d_1 = 8.695,\ d_2 = 6.214\) | M1 |
| \(t = \dfrac{d}{\sqrt{26}} \Rightarrow t_1 = 1.705,\ t_2 = 1.219\) | M1 |
| Interval \(= 0.486\) hrs (29 mins) | A1 |
| (3) |
M1 Find at least one distance
A1 0.49 or better