M4 June 2006 Q3

EdexcelOld spec10 marksVectors

3. A cyclist \(C\) is moving with a constant speed of 10 m s\(^{-1}\) due south. Cyclist \(D\) is moving with a constant speed of 16 m s\(^{-1}\) on a bearing of 240\(^\circ\).

(a) Show that the magnitude of the velocity of \(C\) relative to \(D\) is 14 m s\(^{-1}\). (3)

At 2 pm, \(D\) is 4 km due east of \(C\).

(b) Find

(i) the shortest distance between \(C\) and \(D\) during the subsequent motion,

(ii) the time, to the nearest minute, at which this shortest distance occurs. (7)

Velocity triangle: v_C = 10 due south, minus v_D = 16 at 60 degrees to it, and the relative velocity v_CD; angle theta at the top and alpha opposite the side 10 — Mark scheme (a)

C and D 4 km apart on a horizontal line; relative velocity of C at 8.213 degrees below the line CD; N is the foot of the perpendicular from D — Mark scheme (b)

Scheme	Marks
\(\left\|{}_C\mathbf{v}_D\right\|^2 = 10^2 + 16^2 - 2 \times 10 \times 16\cos 60^\circ\)	M1 A1
\(= 196\)
\(\left\|{}_C\mathbf{v}_D\right\| = 14\) ms\(^{-1}\ \ *\)	A1
	(3)

Scheme	Marks
\(\alpha\) is acute (opposite shortest side)
\(\dfrac{\sin\alpha}{10} = \dfrac{\sin 60^\circ}{14}\)	M1
\(\Rightarrow \alpha = 38.213^\circ\ldots\)	A1

Scheme	Marks
(i) \(\ DN = 4000\sin 8.213\)	M1
\(\simeq 571\) m \(\left(\dfrac{4000}{7}\right)\)	A1
(ii) \(\ t = \dfrac{4000\cos 8.213^\circ}{14}\) secs.	M1 A1
\(\simeq 282.78\ldots\) secs.
Time is 2.05 pm (nearest minute)	A1
	(7)
	(10 marks)

Notes