00:01
For this problem, we are given the height of an object in meters at t seconds defined by s of t, which is equal to negative 4 .9 t squared plus 39 .2t plus 44 .1.
00:15
For the first part, we are to find the velocity of this object at t equals zero.
00:20
Now, the velocity v of t is equal to the derivative of s, so it's s primed t.
00:30
That's the derivative with respect to t of negative 4 .9 t squared plus 39 .2t plus 44 .1.
00:40
That's negative 4 .9 times 2t plus 39 .2, which is equal to negative 9 .8t plus 39 .2.
00:51
At t equals 0, we have velocity equal to negative 9 .8 times 0 plus 39 .2, that's equal to 39 .2 meters per second.
01:07
Next, we want to find the maximum height of this object and when it occurs.
01:15
Now, the maximum height occurs at the point wherein velocity changes sign, that is, from plus to minus.
01:25
So to determine this value of t, the first thing you have to do is to find the value of t wherein v of t is zero.
01:34
From part a, velocity is equal to negative 9 .8t plus 39 .2.
01:40
So, v of t equals 0, that's just negative 9 .8t plus 39 .2 equals 0, gives us t that's equal to negative 39 .2, all over negative 9 .8, that's equal to 4.
02:01
And then you will check if our velocity changes sign at this point from positive to negative.
02:10
By first partitioning our interval from zero to infinity using this value of t.
02:17
So say we have this number line from zero to infinity, and then we have four here.
02:23
So our intervals are from zero to four, and then from four to positive infinity.
02:29
Then we will pick test values from each interval...