00:01
Hi, in this question, given that the position of a particle travelling along a horizontal line i .e.
00:09
S of t equals 2t cube minus 18t square minus 96t plus 5, we need to determine the acceleration of the particle when the velocity is zero.
00:28
So, that first we need to find the velocity v equals ds by dt.
00:34
On differentiating the given equation, then we get 3 twos are 6t square minus 18 into 2 becomes 36t minus 96.
00:45
On differentiating constant term, we get 0.
00:50
And the next we know that the formula a equals dv by dt which is equal to d by dt of 6t square minus 36t minus 96 which is equal to 6 into 2 becomes 12t minus 36.
01:12
Similarly, on differentiating the constant term, we get 0.
01:16
Here, velocity is 0 i .e.
01:20
V equals 0...