(a) P varies directly as Q and inversely as the square of R. If P = 1 when Q = 8 and R = 2, find the value of Q when P = 3 and R = 5. (b) An aeroplane flies...

Answer Details

(a) \(P\propto\dfrac{Q}{R^{2}}\Rightarrow P=\dfrac{kQ}{R^{2}}\).

Using \(P=1,\ Q=8,\ R=2\): \(1=\dfrac{8k}{4}=2k\Rightarrow k=\tfrac{1}{2}\).

When \(P=3,\ R=5\): \(3=\dfrac{\tfrac{1}{2}\,Q}{25}=\dfrac{Q}{50}\Rightarrow Q=150\).

\(Q=150\)

(b)(i) \(A(20^{\circ}N,60^{\circ}E)\) and \(B(20^{\circ}N,20^{\circ}E)\) are on the same parallel of latitude, so the flight is along the parallel \(20^{\circ}N\). Difference in longitude \(=60-20=40^{\circ}\).

\[\text{Distance}=\frac{40}{360}\times2\pi R\cos20^{\circ}=\frac{40}{360}\times2(3.142)(6400)(0.9397)\approx 4199\text{ km}\]\[\text{Average speed}=\frac{4199}{6}\approx 700\text{ km/h (3 s.f.)}\]

(ii) Flying due North from \(B\) is along a meridian, where distance \(=\dfrac{\theta}{360}\times2\pi R\), \(\theta\) the change in latitude.

\[420=\frac{\theta}{360}\times2(3.142)(6400)\Rightarrow \theta=\frac{420\times360}{40217.6}\approx 3.76^{\circ}\]

Starting at \(20^{\circ}N\) and moving north, latitude of \(C=20+3.76\approx 24^{\circ}N\).

Latitude of C \(\approx 24^{\circ}N\).