Thank you to the two dedicated people who have been helping me this evening. Here is my final problem for the night, and here is what I'm thinking.
For A, I think I need to take the derivative of the velocity function to give me the acceleration function. I think the answer is [tex]v'(t) = - \pi cos( \pi t) - 2 \pi - 5[/tex]. Hopefully this first answer is correct.
I get confused at B. I understand that this question is looking for a minimum, so I need to think about critical points and extrema. Do you have to take the derivative of the acceleration function or just use the acceleration function as found in A? I think the former.
C goes back to velocity. Do you use the derivative of the velocity function to solve this (which is actually the acceleration function in A)?

Part A

Yes you find the derivative of v(t) to get the acceleration a(t). Since, v'(t) = a(t)

You are close but not quite there.
The portion -pi*cos(pi*t) is correct, and so is the -5 at the end. The part you have incorrect is the 2pi. It should be positive instead of negative.

Answer: -pi*cos(pi*t)+2pi-5

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Part B

a(t) = -pi*cos(pi*t)+2pi-5
a'(t) = -pi*pi*(-sin(pi*t))+0 ... derivative of cosine is negative sine; use chain rule
a'(t) = (pi)^2*sin(pi*t)

Find the solutions to a ' (t) = 0 on the interval [0,3]
The solutions are t = 0, 1, 2, 3

You'll find that
The local mins are at t = 0 and t = 2
The local maxes are at t = 1 and t = 3
When you use the first derivative test (I'm skipping the work to show this as it is way too lengthy)

Only focus on the local mins t = 0 and t = 2
We'll focus on the endpoints t = 0 and t = 3
It's perfectly possible to have a critical point be an endpoint as long as the function is defined on both sides of the critical point.

So we will plug t = 0, t = 2, t = 3 into the a(t) function to see which is smallest.

Plug in t = 0
a(t) = -pi*cos(pi*t)+2pi-5
a(0) = -pi*cos(pi*0)+2pi-5
a(0) = -1.8584 (approximate)

Plug in t = 2
a(t) = -pi*cos(pi*t)+2pi-5
a(2) = -pi*cos(pi*2)+2pi-5
a(2) = -1.8584 (approximate)

Plug in t = 3
a(t) = -pi*cos(pi*t)+2pi-5
a(3) = -pi*cos(pi*3)+2pi-5
a(3) = 4.4248 (approximate)

The output -1.8584 is smallest.

Answer: -1.8584

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Part C

v(t) = (2pi-5)*t - sin(pi*t)
v ' (t) = a(t) = -pi*cos(pi*t)+2pi-5

Solve a(t) = 0 for t on the interval [0,2]
The solutions are these approximate values
t = 0.3660 and t = 1.6339

You'll find that a local min is at t = 0.3660 while a local max is at t = 1.6339

Focus on the local max. Ignore the local min.
We'll also focus on the endpoints as well

So we'll plug t = 0, t = 1.6339, and t = 2 into the v(t) function to see which is the largest.

Plug in t = 0
v(t) = (2pi-5)*t - sin(pi*t)
v(0) = (2pi-5)*(0) - sin(pi*(0))
v(0) = 0

Plug in t = 1.6339
v(t) = (2pi-5)*t - sin(pi*t)
v(1.6339) = (2pi-5)*(1.6339) - sin(pi*(1.6339))
v(1.6339) = 3.0094 (approximate)

Plug in t = 2
v(t) = (2pi-5)*t - sin(pi*t)
v(2) = (2pi-5)*(2) - sin(pi*(2))
v(2) = 2.5664 (approximate)

The output 3.0094 is the largest, so this is the absolute max (largest velocity) on the interval [0,2]

Answer: 3.0094