An aeroplane flies due west for 3 hours from P (lat. 50°N, long. 60°W) to a point Q at an average speed of 600km/h. The aeroplane then flies due south from ...
Question 1 Report
An aeroplane flies due west for 3 hours from P (lat. 50°N, long. 60°W) to a point Q at an average speed of 600km/h. The aeroplane then flies due south from Q to a point Y 500km away. Calculate, correct to 3 significant figures,
(a) the longitude of Q ;
(b) the latitude of Y . [Take the radius of the earth = 6400km and \(\pi = \frac{22}{7}\)].
Answer Details
Distance flown due west \(=\text{speed}\times\text{time}=600\times3=1800\) km, along the parallel of latitude \(50^{\circ}N\).
(a) Longitude of Q. Distance along a parallel \(=\dfrac{\theta}{360}\times2\pi R\cos(\text{lat})\), where \(\theta\) is the change in longitude.
\[1800=\frac{\theta}{360}\times2\times\frac{22}{7}\times6400\times\cos50^{\circ}\]\(2\times\tfrac{22}{7}\times6400=40228.6\); \(\cos50^{\circ}=0.6428\); product \(=25859\).
\[\theta=\frac{1800\times360}{25859}\approx 25.1^{\circ}\]Flying due west from long. \(60^{\circ}W\) increases the western longitude: \(60+25.1=85.1^{\circ}W\).
Longitude of Q \(\approx 85.1^{\circ}W\) (3 s.f.).
(b) Latitude of Y. Flying due south is along a meridian: distance \(=\dfrac{\phi}{360}\times2\pi R\).
\[500=\frac{\phi}{360}\times40228.6\Rightarrow \phi=\frac{500\times360}{40228.6}\approx 4.47^{\circ}\]Moving south from \(50^{\circ}N\): latitude of \(Y=50-4.47\approx 45.5^{\circ}N\).
Latitude of Y \(\approx 45.5^{\circ}N\) (3 s.f.).
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