Question

A fishing boat leaves port at 12 miles per hour at a bearing of 70∘ for 5 hours, then turns to a bearing of 130∘ at 9 miles per hour for 5 hours, and finally changes to a bearing of 50∘ at 9 miles per hour for 2 hours. At this point, the boat heads directly back to port at a speed of 9 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

I will be happy to critique your work or thinking.

Answer 2

for my final vector, the distance between the final position and the port I did <((60*cos(20))+(45*cos(-40))+(18*cos(40))), ((60*sin(20))+(45*sin(-40))+(18*sin(40)))>
= <104.64,3.17>
(104.64 mi)/(9 mi/hr) = 11.62692857 hours (this is correct)

But i can't get the bearing on the way back.

I tried tan^-1(3.17/104.64), and a dot b = |a||b|cos(x), nothing works

Answer 3

final postion:
E:60sin70deg + 45sin130deg+18sin50deg=92.7
N:60cos70deg + 45cos130deg+18cos50deg=3.16
angle: arc tan E/N= 92.7/3.16 =88.0 deg
so the bearing back has to be 180+88=268 deg
time =distance/speed= (1/9) * sqrt(92.7^2+3.16^2)=10.3 hours
check my work, it does not agree with your "correct" answer.
I dont follow this part of your postion(((((60*sin(20))(((9

Answer 4

Displacement = 60[70o] + 45[130o] + 18[50o].
X = 60*sin70 + 45*sin130 + 18*sin50 = 104.6 mi.
Y = 60*Cos70 + 45*Cos130 + 18*Cos50 = 3.17 mi.

Disp. = sqrt(X^2 + Y^2) = 104.65 mi. = Straight-line distance from starting point.

D = V * T.
a. T = D/V = 104.65/9 = 11.63 h.

b. TanA = X/Y, A = 88.3o CW from +Y-axis(Bearing).