A small boat is headed a harbor 32km directly northwest of its current position when its suddenly engulfed in heavy fog. The captain maintain a compass bearing of northwest and a speed of 10km/h relative to the water.The fog lift 3.o h later and the captain notes that he is now exactly 4.0km south of the harbor.
(a) what was the average velocity of the current during those 3.0h?
(b) In what direction should the boat have been heading to reach its destination along a straight course?
(c) what would its travel time have been if it had followed a straight course?
Sorry i've been trying to solve it but i won't figure it out..
There are a couple of ways to consider this, and I'm probably mixing them, so check my sanity as well as my thinking! I'm not at all sure I've got this right. And draw a small sketch of the points; it makes a whole lot more sense that way.
(a)
Consider the harbor as the origin. Then the boat starts at the point (16sqrt(2), -16sqrt(2)), 32km SE of it.
With no current, the boat would have been at the point (sqrt(2), sqrt(2)) after 3 hours (that's 2km SE of the harbor).
The boat actually ends up at the point (-4,0) after 3 hours.
Subtract the points to get the displacement due to current:
(-sqrt(2), -4+sqrt(2))
That's how far the current carried it offcourse in 3 hours.
Pythagoras will give you the distance; divide by 3 to get the km/h.
(b)
I suppose "a little northerly of where it was actually heading" won't do as an answer? :-) No harm in trying!
OK. The boat needs to subtract the vector of the current to compensate, so it should have been heading:
(-sqrt(10) -sqrt(2)/3, sqrt(10)+(4-sqrt(2))/3)
which I approximate as
(-3.633, 4.024)
to make 10kph relative to the harbor.
(I should be writing this in vector notation, but that's even more awkward to do in ASCII)
The boat can't do that, since it's more than 10kph, but the direction is right. So what's that angle? Consider an RAT with those sides as the adjacent and opposite. The sin of the angle with the x-axis is 4.024 / 5.421, so the angle is about 47.9degrees - that is, 2.9 degrees north of NW.
Aaach. In (a) I meant
"The boat actually ends up at the point (0,-4) after 3 hours. "
Sorry for the confusion.
No problem! Let's break down each part of the problem to find the answers.
(a) To determine the average velocity of the current, we need to consider the displacement of the boat relative to the earth's surface. The boat maintained a compass bearing of northwest, which means it was moving in the northwest direction regardless of the current.
During the 3 hours of heavy fog, the boat traveled a distance of 10 km/h * 3 h = 30 km relative to the water. However, when the fog lifted, the captain observed that the boat was 4.0 km south of the harbor.
Since the boat's displacement due to the current was in the south direction, we can calculate the average velocity of the current as follows:
Average Velocity of Current = Displacement due to Current / Time taken = 4.0 km / 3 h = 1.33 km/h
Therefore, the average velocity of the current during those 3.0 hours was 1.33 km/h towards the south.
(b) To determine the direction the boat should have been heading to reach its destination along a straight course, we need to find the angle between the northwest direction and the direction of the current.
Since the boat's compass bearing was northwest and the current was pushing it slightly south, the resultant direction of the boat's path should have been slightly more towards the west.
To find the exact angle, we can use trigonometry. Let's assume the angle between the original northwest direction and the boat's actual path is θ.
tan(θ) = (Displacement due to Current) / (Displacement relative to Water)
tan(θ) = 4.0 km / 30 km
θ ≈ 7.6 degrees
Therefore, the boat should have been heading approximately 7.6 degrees west of northwest to reach its destination along a straight course.
(c) To determine the travel time if the boat had followed a straight course, we can consider the distance between the boat's final position (4.0 km south of the harbor) and its destination (32 km directly northwest of its original position).
Using Pythagoras' theorem, we can find the straight-line distance between the two points:
Distance = sqrt((Displacement due to Current)^2 + (Displacement relative to Water)^2)
Distance = sqrt(4.0 km^2 + 30 km^2)
Distance ≈ 30.17 km
To calculate the travel time, we can use the boat's speed relative to the water:
Travel Time = Distance / Speed
Travel Time ≈ 30.17 km / 10 km/h
Travel Time ≈ 3.02 hours
Therefore, if the boat had followed a straight course, its travel time would have been approximately 3.02 hours.
I hope this helps! Let me know if you have any further questions.