00:01
Hello, so we're told that our position function is given by x of t, which is equal to 2t times e to the negative t.
00:08
We want to find the acceleration at time t equals zero.
00:12
The acceleration is the second derivative.
00:14
So first we find the velocity, which is going to be the first derivative, so that's going to be x prime of t.
00:21
Okay, so the first derivative here is equal to, well, just two times one times e to negative t, and then minus t.
00:31
Times e to negative t, which is just going to be equal to 2e to the negative t times one minus t.
00:51
Okay, and then we can differentiate here again to find the second derivative.
00:57
So we get that acceleration.
01:01
It's equal to the second derivative.
01:03
The second derivative here is going to be equal to, well, the second derivative, we call that x double prime of t.
01:13
So we differentiate again, and we get 2e to the negative t, and then times 2t minus 3.
01:25
Oops, 2t minus 3 there...