A fishing boat leaves port at 8 miles per hour at a bearing of 220∘ for 5 hours, then turns to a bearing of 270∘ at 11 miles per hour for 4 hours, and finally changes to a bearing of 300∘ at 9 miles per hour for 4 hours. At this point, the boat heads directly back to port at a speed of 4 miles per hour. Find the time it takes the boat to return to port as well as the boat's bearing as it does.

ships sail on headings, not bearings. A bearing is the direction of an object from your position. (It may also be your heading, if you are going directly there.)

So, can you at least convert those times and speeds to distance vectors?

Add them all up x-y wise and figure the final location.

Then just figure the distance and heading. (the bearing of the port)

To find the time it takes for the boat to return to port, we need to break down the journey into different segments and calculate the total time for each segment.

Segment 1: In the first segment, the boat travels for 5 hours at a speed of 8 miles per hour on a bearing of 220∘. We can use the formula distance = speed × time to find the distance traveled in this segment: distance = 8 × 5 = <<8*5=40>>40 miles.

Segment 2: In the second segment, the boat travels for 4 hours at a speed of 11 miles per hour on a bearing of 270∘. Following the same formula, distance = speed × time, we have distance = 11 × 4 = <<11*4=44>>44 miles.

Segment 3: In the third segment, the boat travels for 4 hours at a speed of 9 miles per hour on a bearing of 300∘. Again, using the distance = speed × time formula, we get distance = 9 × 4 = <<9*4=36>>36 miles.

Segment 4: In the final segment, the boat travels at a speed of 4 miles per hour. We don't have a specific time for this segment, but since we know the distances covered in the previous segments, we can calculate the time it takes for the boat to travel this distance. Adding up the distances from the first three segments, we have 40 + 44 + 36 = <<40+44+36=120>>120 miles. To find the time, we use the formula time = distance / speed: time = 120 / 4 = <<120/4=30>>30 hours.

Now, let's find the bearing of the boat as it returns to port. From the last segment, we know that the boat's bearing is 300∘. To return to port, the boat will need to turn around, so the bearing will be the opposite direction, which is 180∘.

Therefore, it takes the boat 30 hours to return to port, and the boat's bearing as it does is 180∘.