M4 June 2006 Q1

EdexcelOld spec5 marksVectors

1. At noon, a boat \(P\) is on a bearing of 120\(^\circ\) from boat \(Q\). Boat \(P\) is moving due east at a constant speed of 12 km h\(^{-1}\). Boat \(Q\) is moving in a straight line with a constant speed of 15 km h\(^{-1}\) on a course to intercept \(P\). Find the direction of motion of \(Q\), giving your answer as a bearing. (5)

Velocity triangle at Q: line QP on bearing 120 degrees; 12 drawn due west (minus v_P), 15 (v_Q) from its end to the line QP, angle 150 degrees, angle alpha between v_Q and QP — Mark scheme

Scheme	Marks
Diagram	M1
\(\dfrac{\sin\alpha}{12} = \dfrac{\sin 150^\circ}{15}\)	M1 A1
\(\Rightarrow \sin\alpha = \dfrac{6}{15}\)
\(\Rightarrow \alpha = 23.6^\circ\)	A1
\(\therefore\) Course is \(096\ (.4^\circ)\)	A1
	(5)

Notes