Posted by **Julie** on Thursday, October 8, 2009 at 1:11pm.

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..

- physics -
**jim**, Thursday, October 8, 2009 at 6:32pm
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.

- physics -
**jim**, Thursday, October 8, 2009 at 6:41pm
Aaach. In (a) I meant

"The boat actually ends up at the point (0,-4) after 3 hours. "

Sorry for the confusion.

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