00:01
In this problem we are given the function s of t is equal to negative 12 t square plus 120 t and this function denotes the position of a particle moving along a line.
00:14
We are asked to find the average velocity in different time intervals.
00:18
The first time interval is 1 .6.
00:21
Now let v average denote the average velocity.
00:24
Then it is the average rate of change of this s of t because velocity is the rate of change.
00:31
Of the displacement function here it is the position function now by definition for a function f of x the average rate of change of the function in an interval closed interval a b is denoted as f average and it is defined as f of b minus f of a divided by b minus a here in this interval the average velocity is the average rate of change of the position function by definition it is s of 6 minus s of 1 by so 6 minus 1.
01:02
Now s of 6 is obtained by substituting in s of t, the value t is equal to 6.
01:07
So that is negative 12 times 6 square plus 120 times 6 which is negative 432 plus 720 plus 720 which is 288.
01:24
So that s of 6 is 288.
01:26
Similarly s of 1 is negative 12 plus 120.
01:30
That is when obtained it is obtained when substituting t is equal to 1 in s of t and that is 1 .8 so that the average velocity during this time interval is 288 minus 108 divided by 5 which is 180 by 5 upon simplification this is 36 so that the average velocity during this time interval is 36.
01:53
The second time interval is interval 15 and in this interval the average velocity is is calculated as the average rate of change of the position function that is s of 5 minus s of 1 by 5 minus 1...