A skater travels in a straight line 8.5*10^2m[25 degress N of E] and then 5.6810^2m in a straight line [21 degrees E of N]. The entire motion takes 4.2 min. What is the skater's displacement angle?

Question

istim · Answer

Figured out the displacement. 1.4*10^3m. The angle's iffy.

jamesj · Answer

The amount of time taken is irrelevant.  For each leg of the journey figure out the angle of displacement.  Call each vector, $ v_1 $ and $ v_2 $.  Each vector has an x and a y component.  Write
$$ v_1 = (x_1,y_1), v_2 = (x_2,y_2) $$
Then at the end the skater is at $ v_1 + v_2 = (x_1 + x_2, y_1 + y_2 ) $ and can calculate her angle of displacement by taking the tan of the ratio of the y displacement over the x displacement.

It is identical in principle to the last problem you had.

istim · Answer

It is the last problem I had. I just didn't ask for the angle that time. Reposted...
Anyways, I usedA=sin^-1(asinB/b). Does that work?

jamesj · Answer

You should also draw a diagram so you can see at least directionally where the answer is.  No, use tan(y/x)

istim · Answer

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