00:01
In this problem, we need to determine the acceleration of an object at a given time t, but the displacement s is given as 26t minus 4 .9t square, where t represents the time.
00:16
Now, the acceleration of an object is the second derivative of s.
00:21
So let us determine the second derivative.
00:24
First of all, let us determine the first derivative.
00:27
That will be the derivative of 26t, which will be 26 times the derivative of t which is 1, minus 4 .9 times the derivative of t squared, which is 2t.
00:37
So we get 26 minus 4 .9 times 2, which is equal to 9 .8 times t.
00:46
Next, let us determine the second derivative of the function.
00:54
So that will be equal to the derivative of 26, which is 0 because 26 is a constant, minus 9 .8 times the derivative of t which is equal to 1...