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