M4 June 2012 Q2
2. A ship \(A\) is moving at a constant speed of 8 km h\(^{-1}\) on a bearing of \(150^\circ\). At noon a second ship \(B\) is 6 km from \(A\), on a bearing of \(210^\circ\). Ship \(B\) is moving due east at a constant speed. At a later time, \(B\) is \(2\sqrt{3}\) km due south of \(A\).
Find
| Scheme | Marks |
|---|---|
| With \(B\) as origin, | |
| \(\mathbf{r}_A = (6\sin 30\,\mathbf{i} + 6\cos 30\,\mathbf{j})\) | M1 |
| \(= (3)\mathbf{i} + (3\sqrt{3})\mathbf{j}\) | A1 |
| \(\mathbf{r}_B = vt\mathbf{i}\) or \(\mathbf{v}_B = v\mathbf{i}\) | B1 |
| \((v - 4)\mathbf{i} + (4\sqrt{3})\mathbf{j}\) | M1 |
| or \((v - 8\sin 30)\mathrm{i} + (8\cos 30)\mathrm{j}\) | A1 |
| When \(B\) is \(2\sqrt{3}\) km south of \(A\), | |
| \(-3\sqrt{3} + 4\sqrt{3}t = -2\sqrt{3} \Rightarrow t = \tfrac{1}{4}\) | M1 A1 |
| \(vt - 3 - 4t = 0 \Rightarrow v = 16\) | M1 A1 |
| When \(B\) is due east of \(A\), | |
| \(-3\sqrt{3} + 4\sqrt{3}t = 0 \Rightarrow t = \tfrac{3}{4}\) i.e. at 12.45 pm | M1 A1 |
| then distance \(AB = 16 \times \tfrac{3}{4} - 3 - 4 \times \tfrac{3}{4} = 6\) km. | M1 A1 |
| (13 marks) |
Notes
M1 Express the original relative positions in component (vector) form – one term correct.
A1 Both terms correct (substitution of trig values not required).
B1 Position of \(B\) at time \(t\) (seen or implied)
M1 Express the relative velocity in component form – one term correct.
A1 Both terms correct
M1 Compare \(\mathbf{j}\) displacement with \(\pm 2\sqrt{3}\) and solve for \(t\)
A1 cao
M1 Equate \(\mathbf{i}\) displacement to zero and substitute their value of \(t\).
A1 cao
M1 Equate \(\mathbf{j}\) displacement to zero and solve for \(t\).
A1 Any equivalent form for the time.
M1 Substitute their \(v\) & \(t\) in the \(\mathbf{i}\) displacement and evaluate
A1 cao. Must be a scalar.
Alternative
| Triangle \(ABC\): cosine rule gives \(BC^2 = 36 + 12 - 2 \times 6 \times 2\sqrt{3}\cos 30\) | M1 A1 |
| Solve for \(BC\) and \(\angle ABC\) \(BC = 2\sqrt{3}\), \(\rightarrow\) triangle is isosceles | M1 A1 |
| \(\angle B\) in velocity triangle is \(30^\circ\) | B1 |
| Trig in rt\(\angle\) triangle gives relative velocity \(= 8 \times \tan 60 = 8\sqrt{3}\) | M1 A1 |
| \(\angle\mathrm{APB} = 30^\circ\) (angles of a triangle) so triangle is isosceles and distance \(AP = 6\)km | M1 A1 |
| Using cosine rule or symmetry of isosceles triangle, distance \(BP = 6\sqrt{3}\) | M1 A1 |
| Time taken \(= \frac{6\sqrt{3}}{8\sqrt{3}} = \frac{3}{4}\) hr, time is now 12.45 | M1 A1 |
The given information provides us with two triangles - velocities in bold. Fix \(A\) and \(B\) follows the path \(BP\). \(C\) is the point when \(B\) is due South of \(A\), and \(P\) when it is due East.
