A ship leaves an island (5^o, 45^oE) and sails due east for 120 hours to another island. The average speed of the ship is 27 knots

A ship leaves an island (5^o, 45^oE) and sails due east for 120 hours to another island. The average speed of the ship is 27 knots
a) Calculate the distance between the two islands
i) in nautical miles
ii) in kilometres
b) Calculate the speed of the ship in kilometres per hour
c) Find the position of the second island. (Take 1 nm to be 1.853 km and the radius of the earth to be 6370 km)

Answers

i) Distance in nautical miles

Distance

= speed x times

= 27 nm/hr x 120 hrs

= 3240 nm

b) In km

1 nm = 1.853 km
3 240 nm= ?

= 3240 x 1.853

= 6 003.72 km

= 6 004 km

b) Speed in kilometres per hour

= $\frac{Distance}{Time}$ = $\frac{6 003.72}{120}$

= 50.03 km/hr

c) A (5^oN, 45^oE), B(5^oN, x^oE) let

longitude difference be x

= $\frac{x}{360}$ x 2 x 3.142 x 6370 x cos5^o

= 5 994

= $\frac{x}{360}$ x 2 x 3.142 x 6 370 x 0.996 = 5994

110.6x = 5 994; x = 5994

110.6x = 5 994; x = 54.1

longitude of B

= 99.1^oE

position of B

= (5^oN,99.1^oE)

---

johnmulu answered the question on March 8, 2017 at 06:19

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