A ship sails from point A 40 circ N 28 circ E t... - SS3 Mathematics Longitude and Latitude Question
A ship sails from point \(A(40{^\circ}N,28{^\circ}E)\) to point \(B(40{^\circ}N,\ 25{^\circ}W)\) and then to point \(C(15{^\circ}S,\ 25{^\circ}W)\). Take the radius of the earth, \(R = 6400\ km\) and \(\pi = 3.142\). Calculate:
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The distance from \(A\) to \(B\) and then to \(C\) in nautical miles (\(1\ nautical\ mile\ = \ 1.86km\))
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The average speed of the ship in knots if the whole journey takes \(49\) hours (\(1\ nautical\ mile\ per\ hour = 1\ knot\))
note: nautical miles and\ knots are units associated with sea and air travel
Distance from Point \(A\ \)to \(B\ \)is along a small circle, latitude \(40{^\circ}N\), thus:
\(\alpha = 28{^\circ} + 25{^\circ} = 53{^\circ}\)
\[R = 6,400km\]
\[\pi = 3.142\]
\[\theta = 40{^\circ}\]
\[d = \frac{53}{360} \times 2 \times 3.142 \times 6400{\times cos}{40{^\circ}} = 4,535.69\ km\]
Distance from Point \(B\ \)to \(C\ \)is along a great circle, longitude \(25{^\circ}W\), thus:
\(Angular\ difference\ = \ 40{^\circ} + 15{^\circ} = 55{^\circ}\)
\[Distance\ between\ the\ points = \frac{\theta}{360} \times 2\pi R\]
\[Distance\ between\ the\ points = \frac{55}{360} \times \frac{2}{1} \times \frac{22}{7} \times \frac{6400}{1} = 6,146.03\ km\]
a. Total distance from point \(A\) to \(B\) to \(C\) = \(4,535.69\ km + \ 6,146.03\ km = 10,681.72\ km\)
Total distance in nautical miles = \(\frac{10,681.72}{1.86} = 5,742.86\ nautical\ miles\)
b. Average speed = \(\frac{distance}{time} = \frac{5,742.86}{49} = 117.20\ knots\) (\(1\ nautical\ mile\ per\ hour = 1\ knot\))
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