M4 June 2017 Q5

EdexcelOld spec9 marksVectors

5. A cyclist riding due north at a steady speed of 12 km h\(^{-1}\) notices that the wind appears to come from the north-west. At the same time, another cyclist, moving on a bearing of \(120^\circ\) and also riding at a steady speed of 12 km h\(^{-1}\), notices that the wind appears to come from due south. The velocity of the wind is assumed to be constant.

Find

(i) the wind speed,

(ii) the direction from which the wind is blowing, giving your answer as a bearing. (9)

Diagram: two velocity triangles sharing the wind velocity w; 12 north with x at 45 degrees, and 12 on bearing 120 with y vertical, angle theta between 12 and w — Mark scheme

Scheme	Marks
	B1 B1
\(12\cos 30^\circ = x\cos 45^\circ\)	M1
\(x = 12\sqrt{\dfrac{3}{2}}\ \left(= 6\sqrt{6} = 14.6\ldots\ldots\right)\)	A1
\(w^2 = 12^2 + x^2 - 2\times 12x\cos 45^\circ\)	M1
\(w = 10.5\) (km h\(^{-1}\))	A1
EITHER: Sine Rule: \(\dfrac{\sin\theta}{x} = \dfrac{\sin 45^\circ}{w}\) \(\left(\dfrac{\sin\theta}{12} = \dfrac{\sin 60^\circ}{w}\right)\)	M1
\(\theta = 81.2^\circ\)	A1
Direction \(261^\circ\)	A1
OR: \(\updownarrow\ w\cos\theta = 12 - 12\cos 30^\circ\) \(\leftrightarrow\ w\sin\theta = 12\cos 30^\circ\)
\(\Rightarrow \tan\theta = \dfrac{\cos 30^\circ}{1 - \cos 30^\circ},\)	(M1)
\(\theta = 81.2^\circ\)	(A1)
Direction \(261^\circ\)	(A1)
	(9)
	(9 marks)

\(\mathbf{w} = \begin{pmatrix}x\cos 45^\circ\\12 - x\cos 45^\circ\end{pmatrix}\)	B1
\(\mathbf{w} = \begin{pmatrix}12\cos 30^\circ\\-12\sin 30^\circ + y\end{pmatrix}\)	B1
\(12\cos 30^\circ = x\cos 45^\circ\)	M1
\(x\cos 45^\circ = 6\sqrt{3}\)	A1
\(\|\mathbf{w}\|^2 = 3\times 36 + \left(12 - 6\sqrt{3}\right)^2\)	M1
\(\|\mathbf{w}\| = 10.5\) (km h\(^{-1}\))	A1
\(\tan\theta = \dfrac{6\sqrt{3}}{12 - 6\sqrt{3}}\)	M1
\(\theta = 81.2^\circ\)	A1
Direction \(261^\circ\)	A1
	(9)

M4 June 2017 Q5

M4 June 2017 Q5

Notes

5 alt