00:01
Yes, hello.
00:01
So the average velocity over some time interval from a to b is just going to be equal to, so we have a function is going to be, again, s of t, and we're looking at the average velocity over a to b.
00:16
That's going to be s of b.
00:18
So we evaluate the function at b.
00:19
We subtract off the function evaluated at a.
00:23
So s of b minus s of a, all divided by b minus a.
00:28
So to solve this problem for these intervals, we have to just extract the position values, s of a and s of b, from the graph.
00:37
So we have, one for part a, so we have that s of 0 .5 is about 6, s of 1 is about 5, s of 1 .5 is about 4 .2, s of 2 is about 4, and s of 2 .5 is about 3 .2...