00:01
In this question, given the position function of a particle or an object, we are asked to determine particles acceleration and velocity, and then determine the intervals where the particle is speeding up and slowing down.
00:17
First, recall that the velocity function equals to the derivative of the position function.
00:28
So it's the derivative of 2t cube minus 3 t squared minus, 36 t plus 5.
00:37
The derivative of 2t cube equals to 6 times t squared, the derivative of 3 t squared equals to 6 t and the derivative of 36t equals to 36.
00:51
All right, this means that v of t equals to 6 t squared minus 6 t minus 36.
00:59
Now let's find the acceleration function.
01:03
A of t equals to the derivative of v and the derivative of v equals to 12t minus 6.
01:17
All right, we found the velocity function and we found the object's acceleration.
01:27
Now let's find the intervals or which it's speeding up and slowing down...