00:01
Okay, so we have an object that is going along the path, s of t, is equal to t squared, where t is in seconds and s is in feet.
00:13
So s is in feet and t is in seconds.
00:18
All right, so we need to find the average velocity between t equals 2 and t equals 5, right? so the average velocity is going to be s of 5 minus s of 2 divided by 5 minus 2.
00:37
Okay.
00:39
So if we plug in 5 squared minus 2 squared over 3, so 25 minus 4 divided by 3, we get 21, divided by 3, which is equal to 7, and this is in feet per second.
01:01
Now for part b, it says approximate the instantaneous velocity by using smaller and smaller intervals.
01:10
Alright, so first we're going to do, so go from t equals 2 to let's say t equals 2 .1.
01:24
Okay, so if we do that, so if we put again 2 .01 minus s of 2 .01.
01:33
S of 2, divided by 2 .01, minus 2.
01:38
Okay, so that's 2 .01 squared, minus 2 squared, all over 0 .01.
01:48
Now, all right, we get 4 .0401 minus 4, all over 0 .01...