00:01
For this problem, we're given the acceleration function a of t, which is equal to t plus five rates of power of three.
00:07
We're also given the initial velocity, which is equal to four, and the initial position, which is equal to three, since the problem stated that the initial position must be three units to the right of the origin.
00:23
Now, to find v of t, we have to note that v of t is equal to the antiderivative, of the acceleration function, a of t, d t.
00:34
So that's taking the antiderivative of t plus five raise to the third power, and then d t.
00:42
And then we note that if we have antiderivative of x plus a raise the power of n d x.
00:52
This is just x plus a raised to n plus 1 over n plus 1.
00:59
So then this is just t plus 5 raise of power of 4 over 4, but then you have to add a constant c because when you take the derivative of a constant, that's just 0.
01:14
And then to find the value of c, we have to note that the initial velocity is equal to 4...