00:01
Calculate the total vector displacement.
00:04
So from our position o to position a, oa is equal to our v1 times our time in the negative j hat direction.
00:15
So that's equal to 22 times 3 times 60 times negative j hat or negative 3960 j hat.
00:25
From a to b, we have again v2 times t2 times negative i hat.
00:30
So we have 25 times 2 .6 times 60 times negative i hat, which is negative 3 ,900 i hat.
00:40
And from b to c, we have v3 times t3 times sine of 45 degrees times negative i hat, plus cosine 45 degrees times j hat.
00:53
So this turns out to be 30 times 1 times 60 times negative 1 over root 2.
01:01
I -hat plus 1 over root 2 j hat.
01:05
So bc is negative 1272 .7 i hat plus 1272 .7 j hat.
01:14
So our total vector displacement is equal to oa plus ab plus bc which is equal to negative 5172 .7 i hat minus 2687 j hat.
01:30
Then we'd like to find the magnitude of the direction, so our direction, our angle with the direction of our y -axis is equal to inverse tangent of negative 2687, divided by negative 5172 .7, which is 207 .4 degrees.
01:51
And so if we have direction south of west, our direction is 207 .4 minus 180, which is 27 .4 .2 .8, which is 27 .4...