M4 June 2007 Q4
4. At 12 noon, ship \(A\) is 20 km from ship \(B\), on a bearing of 300\(^\circ\). Ship \(A\) is moving at a constant speed of 15 km h\(^{-1}\) on a bearing of 070\(^\circ\). Ship \(B\) moves in a straight line with constant speed \(V\) km h\(^{-1}\) and intercepts \(A\).
It is now given that \(V = 13\).
| Scheme | Marks |
|---|---|
| Fix A | |
| \(v_{\min} = 15\sin 50^\circ\) | M1A1 |
| \(= 11.5\) km h\(^{-1}\) (3 s.f.) | A1 |
| (3) |
Notes
or: triangle without the right angle identified and \(\dfrac{15}{\sin\theta} = \dfrac{v_B}{\sin 50}\)
\(\Rightarrow v_B = \dfrac{15\sin 50}{\sin\theta}\)
minimum value \(\Rightarrow \theta = 90\) for M1
As above for A1A1
M1 Velocity of B relative to A is in the direction of the line joining AB. Minimum V requires a right angled triangle. Convincing attempt to find the correct side.
A1 15 x sin(their \(50^0\))
A1 Q specifies 3sf, so 11.5 only
| Scheme | Marks |
|---|---|
| Ambiguous Sine Rule: 2 possible solutions for \(\alpha\) | B1B1 |
| (2) |
Notes
B1B1 Convincing argument
B1B0 Argument with some merit
| Scheme | Marks |
|---|---|
| \(\dfrac{\sin\alpha}{15} = \dfrac{\sin 50}{13}\) | M1A1 |
| \(\alpha = 62.1^o\) (or \(118^o\)) (smaller value gives larger relative velocity) | A1 |
| \(\Rightarrow\) either \(v = 13\cos 62.1 + 15\cos 50 = 15.72\,kmh^{-1}\) | M1A1 |
| Time \(= \dfrac{20}{their\ 15.72\ldots..}\) | M1 A1 |
| \(= 1.272\ldots\ldots\) hrs | |
| Earliest time is 13.16hrs or 13.17 hrs accept 1.16 (pm) or 1.17 (pm) | A1 |
| (8) | |
| (13 marks) |
Notes
Alternative
| Or \(v^2 = 15^2 + 13^2 - 390\cos 67.9 = 247.27\) | M1 A1 |
| \(v = 15.7\,kmh^{-1}\) |
M1 Use of Sine Rule
A1 Correct expression
A1 (2 possible values,) pick the correct value.
M1 Use trig. to form an equation in v
A1 correct equation
M1 \(time = \dfrac{distance}{speed}\)
A1ft correct expression with their v (not necessarily evaluated)
A1 correct time in hours & minutes
Or: M1 Use of cosine rule
A1 \(13^2 = 15^2 + v^2 - 2 \times 15 \times v \times \cos 50\)
A1 (Award after the next two marks) 15.72 or awrt 15.72
M1 Attempt to solve the equation for \(v\)
A1 \(\dfrac{30\cos 50 \pm \sqrt{(30\cos 50)^2 - 4 \times 56}}{2}\qquad\) (15.72 or 3.562)
Finish as above
(Corrected from the printed mark scheme: the angle is printed as 62,1\(^o\).)