Question

A ship leaves a harbor at noon and sails S50*W at 18 nautical miles per hour until 2:30pm. At that time it changes course and sails N20*W at a reduced speed of 13 nautical miles per hour until 4:00pm, what is the distance of the ship from the harbor at 4:00pm? And what is the ships bearing from the harbor at 4:00pm?

Answer 1

1 nautical miles per hour = 0.514444444 meters per second

Part 1
v =18 nautical mph = 9.26m/s
t = 2.5 h = 9000 s

d = vt
= 83340 m

Part 2
v = 13 nautical mph = 6.69 m/s
t = 1.5 h = 5400 s

d = vt
= 36126m

Part 3
Find distance in the x and y directions,

for part 1, y = 83340 sin 50 = 63842.14 m (apply -ve because of direction if required)
x = 83340 cos 50 = 53569.92 m (apply -ve because of direction if required)

for part 2, y = 36126 sin 20 = 12355.82 m (apply -ve because of direction if required)
x = 36126 cos 20 = 33947.32 m (apply -ve because of direction if required)

Adding the x values and getting the resulatant,

y = -63842.14 - 12355.82
= -76197 m

x = -53569.92 + 33947.32
= -19622.60 m

Resultant = sqrt [(x)^2 + (y)^2]
= 78683.09 m

tan x = 76197 / 19622.60 (-ve signs cancel out)
hence x = 75.56 degrees

Since both x and y were negative overall,

the angle is W 75.56 S

The ship was 78.683 km = 42.485 nautical miles [ W 76 S]