00:01
In this problem we are given with a velocity equation which is v of t is equal to 32 minus 2 t feet per second.
00:12
In the first part of the problem we are as to find the time t for which the object comes to stop.
00:20
That is we need to find the value of t when velocity is equal to zero.
00:25
So let's equate this equation to zero.
00:29
So we will get 32 minus 2 t is equal to 0 from which what we can say is 2 t is equal to 32 and from here we can get the value of t as 32 divided by 2 which is equal to 16 seconds.
00:47
So what we can conclude is when the time is 16 seconds the object comes to stop.
00:55
Now let's move on to the part b in part b we, we, we will.
00:59
Are as to find the distance traveled by the car during the time we found in path a.
01:08
That is, we need to find the value of s when t is equal to 16 seconds.
01:15
We know that velocity is the derivative of distance with respect to the variable t.
01:23
So if we want a value of s, we need to integrate this expression.
01:27
So to find the value of s we need to integrate v of t with respect to t.
01:35
So if we integrate 32 minus 2 t with respect to 32 here and we are integrating with respect to t if we do that we will get 32 times t minus 2 t squared divided by 2 plus c where c is the integration constant.
01:56
Now to find the value of this constant, we can substitute s equal to 0 when t is equal to 0.
02:03
That is, we know that when time is 0 second, this position of the object will not be changed.
02:10
That is, the distance traveled by the object will be 0.
02:14
So, if we substitute s is equal to 0 and t is equal to 0 in this expression, we will get 0 is equal to 32 times 0 minus 0.
02:26
Plus c which implies that the value of c is equal to 0...