00:01
In this problem we're given that s of t, which is equal to negative 16 t squared plus v .0 times t, plus s.
00:07
S .0 is the position function after t seconds of an object, where v sub zero is the initial velocity and s of zero is the initial position.
00:16
We suppose a coin is dropped from a building that is 1 ,296 feet tall.
00:22
We are to find first the position and velocity functions of this coin.
00:30
So for s of t, this is equal to negative 16 t squared plus.
00:37
Initial velocity here, since the coin is dropped from a building would be 0 times t plus s of 0, which is 1 ,296.
00:47
So this is equal to negative 16 t squared plus 1 ,296.
00:53
And for the velocity, v of t is just s prime of t, and that's equal to negative 32t.
01:01
For part b, we are to determine the average velocity on the interval 1 to 2.
01:06
So average velocity, v of t is equal to s of 2 minus s of 1, all over 2 minus 1.
01:16
That's negative 32, rather negative 16 times 2 squared.
01:24
Plus 1 ,296 minus if negative 16 times the square of 1 plus 1 296 all over 1, that's equal to negative 48 feet per second...