M1 June 2006 Q7
7. [In this question the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are due east and north respectively.]
A ship \(S\) is moving with constant velocity \((-2.5\mathbf{i} + 6\mathbf{j})\) km h\(^{-1}\). At time 1200, the position vector of \(S\) relative to a fixed origin \(O\) is \((16\mathbf{i} + 5\mathbf{j})\) km. Find
The ship is heading directly towards a submerged rock \(R\). A radar tracking station calculates that, if \(S\) continues on the same course with the same speed, it will hit \(R\) at the time 1500.
The tracking station warns the ship’s captain of the situation. The captain maintains \(S\) on its course with the same speed until the time is 1400. He then changes course so that \(S\) moves due north at a constant speed of 5 km h\(^{-1}\). Assuming that \(S\) continues to move with this new constant velocity, find
| Scheme | Marks |
|---|---|
| Speed \(= \sqrt{(2.5^2 + 6^2)} = 6.5\) km h\(^{-1}\) | M1 A1 |
| (2) |
Notes
(a) M1 needs square, add and \(\sqrt{\ }\) correct components
| Scheme | Marks |
|---|---|
| Bearing \(= 360 - \arctan(2.5/6) \approx 337\) | M1 A1 |
| (2) |
Notes
(b) M1 for finding acute angle \(= \arctan(2.5/6)\) or \(\arctan(6/2.5)\) (i.e. 67\(^\circ\)/23\(^\circ\)).
Accept answer as AWRT 337.
| Scheme | Marks |
|---|---|
| \(\mathbf{R} = (16 - 3 \times 2.5)\mathbf{i} + (5 + 3 \times 6)\mathbf{j}\) | M1 |
| \(= 8.5\mathbf{i} + 23\mathbf{j}\) | A1 |
| (2) |
Notes
(c) M1 needs non-zero initial p.v. used + ‘their 3’ \(\times\) velocity vector
| Scheme | Marks |
|---|---|
| At 1400 \(\mathbf{s} = 11\mathbf{i} + 17\mathbf{j}\) | M1 A1 |
| At time \(t\), \(\mathbf{s} = 11\mathbf{i} + (17 + 5t)\mathbf{j}\) | M1 A1 |
| (4) |
Notes
(d) Allow 1st M1 even if non-zero initial p.v. not used here
| Scheme | Marks |
|---|---|
| East of \(R \;\Rightarrow\; 17 + 5t = 23\) | M1 |
| \(\Rightarrow t = 6/5 \;\Rightarrow\;\) 1512 hours | A1 |
| (2) |
Notes
(e) A1 is for answer as a time of the day
| Scheme | Marks |
|---|---|
| At 1600 \(\mathbf{s} = 11\mathbf{i} + 27\mathbf{j}\) \(\mathbf{s} - \mathbf{r} = 2.5\mathbf{i} + 4\mathbf{j}\) | M1 |
| Distance \(= \sqrt{(2.5^2 + 4^2)} \approx 4.72\) km | M1 A1 |
| (3) | |
| (15 marks) |
Notes
(f) 1st M1 for using \(t = 2\) or 4 (but not 200, 400, 6, 16 etc) and forming \(\mathbf{s} - \mathbf{r}\) or \(\mathbf{r} - \mathbf{s}\)