An aeroplane flies due North from a town T on the equator at a speed of 950km per hour for 4 hours to another town P. It then flies eastwards to town Q on l...
Question 1 Report
An aeroplane flies due North from a town T on the equator at a speed of 950km per hour for 4 hours to another town P. It then flies eastwards to town Q on longitude 65°E. If the longitude of T is 15°E,
(a) represent this information in a diagram ;
(b) calculate the : (i) latitude of P, correct to the nearest degree ; (ii) distance between P and Q, correct to four significant figures. [Take \(\pi = \frac{22}{7}\); Radius of the earth = 6400km].
Answer Details
(a) Draw a circle for the earth with the equator horizontal and the North pole on top. Mark \(T\) on the equator (longitude \(15^\circ E\)). The plane flies due North along the \(15^\circ E\) meridian to \(P\), then East along the parallel of latitude through \(P\) to \(Q\) on longitude \(65^\circ E\).
(b)(i) Latitude of P: Distance North \(= 950\times 4 = 3800\,\text{km}\), an arc along a meridian: \[3800 = \frac{\theta}{360}\times 2\pi R = \frac{\theta}{360}\times 2\times\frac{22}{7}\times 6400.\] \[2\pi R = 40228.6\,\text{km} \Rightarrow \theta = \frac{3800\times 360}{40228.6} = 34.0^\circ.\] Latitude of \(P \approx \mathbf{34^\circ N}.\)
(b)(ii) Distance P to Q: Along the parallel of latitude \(34^\circ N\), the difference in longitude is \(65^\circ - 15^\circ = 50^\circ\): \[PQ = \frac{50}{360}\times 2\pi R\cos34^\circ = \frac{50}{360}\times 40228.6\times 0.8290.\] \[PQ = 5587.3\times 0.8290 = 4632\,\text{km}.\] Correct to four significant figures, \(|PQ| \approx \mathbf{4632\,\text{km}}.\)
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