Question

A fishing boat leaves port at 5 miles per hour at a bearing of 350∘ for 2 hours, then turns to a bearing of 40∘ at 12 miles per hour for 2 hours, and finally changes to a bearing of 180∘ at 10 miles per hour for 5 hours. At this point, the boat heads directly back to port at a speed of 3 miles per hour. Find the time it takes the boat to return to port as well as the boat's bearing as it does.

Answer 1

d1 = 5mi/h[350o] * 2h = 10mi[350o]

d2 = 12mi/h[40o] * 2h = 24mi[40o]

d3 = 10mi/h[180] * 5h = 50mi[180]

X = 10*Cos350 + 24*Cos40 + 50*Cos180 =
-21.8 Miles.
Y = 10*sin350 + 24*sin40 + 50*sin180 =
13.7 Miles.

d = d1+d2+d3 = X + Yi=-21.8 + 13.7i,Q2.

Tan Ar = Y/X = 13.7/-21.8 = -0.62800
Ar = -32.13o = Reference angle.
A = -32.13 + 180 = 147.9o, CCW.

d = X/cosA = -21.8/Cos147.9 = 25.7 Miles
[147.9o]

T = d/V = 25.7mi[147.9o]/3 = 8.57 Hours.
Direction = 147.9 Degrees.