00:01
Problem number 99 is a position velocity acceleration problem.
00:06
So we're reminded to use this formula equation for an object in free fall, where b sub -0 represents the initial velocity of the object, then s -sub -0 represents the initial position of the object.
00:21
So we're given the initial position of 1360 feet.
00:33
And since the object is just dropped, we'll have initial velocity of zero.
00:39
So we can create the part a has to write the velocity function.
00:45
So v of t or sorry, asks us to first write the position function.
00:51
So s of t would be negative 16t squared plus 1362.
00:59
So then finding the velocity of that object, the velocity is the derivative of acceleration so we would differentiate this using power rule to get negative 32t for part b we're determining the average velocity on the interval from 1 to 2 so to do that we'll need to find the velocity at 1 which is 2 and the velocity at 2 which is negative 64 so velocity would be change in velocity over a change in time.
02:03
So negative 64 minus negative 32, and the change in time is 1.
02:12
So part b asks us to find the average velocity on the interval from 1 to 2.
02:21
So the average velocity is the change in position over a change in time.
02:27
So we would need to find s of 2...