M4 June 2005 Q6

EdexcelOld spec12 marksVectors

6. A ship \(A\) has maximum speed 30 km h\(^{-1}\). At time \(t = 0\), \(A\) is 70 km due west of \(B\) which is moving at a constant speed of 36 km h\(^{-1}\) on a bearing of 300\(^\circ\). Ship \(A\) moves on a straight course at a constant speed and intercepts \(B\). The course of \(A\) makes an angle \(\theta\) with due north.

(a) Show that \(-\arctan\tfrac{4}{3} \leqslant \theta \leqslant \arctan\tfrac{4}{3}\). (7)

(b) Find the least time for \(A\) to intercept \(B\). (5)

A and B 70 km apart on a west-east line; from A a vector of 36 at 30 degrees below the line to A prime; from A prime two vectors of 30 to P and Q on the line, each at angle theta to the north direction and 30 degrees to the line — Mark scheme (a)

Scheme	Marks
Minimum speed for interception \(= 36\sin 30^\circ = 18\)	M1 A1
\(\cos\theta = \dfrac{18}{30}\ \left(= \tfrac{3}{5}\right)\)	M1 A1
\(\Rightarrow \tan\theta = 4/3\)	A1
Explanation	M1 A1 cso
	(7)

Scheme	Marks
\(AQ = 36\cos 30^\circ + 30\sin\theta\)	M1 A2
\(\left(18\sqrt{3} + 24\right)\)
Time \(= \dfrac{70}{\left(18\sqrt{3} + 24\right)} = 1.27\) hrs.	M1 A1
	(5)
	(12 marks)

Notes