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.
I will be happy to critique your work or thinking.
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
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
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).
To find the time it takes for the boat to return to port, we need to calculate the total distance the boat has traveled. The boat travels at a speed of 12 miles per hour for 5 hours, then at a speed of 9 miles per hour for 5 hours, followed by a speed of 9 miles per hour for 2 hours. Finally, it returns to port at a speed of 9 miles per hour.
First, let's calculate the distance traveled during each leg of the journey:
Distance of the first leg = Speed x Time = 12 mph x 5 hours = 60 miles
Distance of the second leg = Speed x Time = 9 mph x 5 hours = 45 miles
Distance of the third leg = Speed x Time = 9 mph x 2 hours = 18 miles
Now, we can find the total distance traveled by summing up all the distances:
Total distance = Distance of first leg + Distance of second leg + Distance of third leg = 60 miles + 45 miles + 18 miles = 123 miles
Next, we need to determine the bearing of the boat as it returns to the port. We know that the boat started at a bearing of 70∘, turned to 130∘, and then to 50∘. To determine the bearing as it returns to port, we need to find the total change in bearing.
The total change in bearing = Initial bearing - Final bearing = 50∘ - 70∘ = -20∘
Since the boat is returning to the port, the final bearing would be the opposite of the initial bearing. So, the final bearing would be 180∘ plus the initial bearing.
Final bearing = 180∘ + Initial bearing = 180∘ + 50∘ = 230∘
Therefore, the boat's bearing as it returns to port is 230∘.
To find the time it takes for the boat to return to port, we can divide the total distance traveled by the speed at which the boat is returning to port:
Time taken to return = Total distance / Speed = 123 miles / 9 mph ≈ 13.67 hours
So, it takes approximately 13.67 hours for the boat to return to port.