Question: ***Disclaimer: I am not a student trying to get free internet homework help. I am an adult who is learning Calculus from a textbook. I am deeply grateful to the members of this community for their time.***

When graphing acceleration or v'(t), should you put a "hole" in the endpoints of the graph?

Let's look at the typical graph of a rock being thrown, where s(t) = inverted parabola. 

v(t) is a diagonal line going from positive to negative (eg: y=-32t+4)

a(t) = v'(t) = constant (negative) (eg: y=-32)

Let's say the rock hits the ground at t=10. What should acceleration be EXACTLY at t=10? Is it still -6, like the times before it? Or does it not exist since at t=10, the rock has stopped. Can you have acceleration when velocity = 0? I guess you can b/c acceleration at the peak position (vertex of parabola) is still negative -6, even though the rock is not moving in that peak's instant. Can you take the derivative of the point at t=10 on the velocity graph? 

I saw the sample problem does NOT create a hole at the very end of the A(t) graph when the rock hits the ground. So, the graph is saying the Acc. is still the same as when it was falling a moment before. It also fills in the A(t) at t=0, even though the rock is not yet moving and has velocity = 0 at t=0

etc etc etc....

Answer: 
This is not a mathematics question, but a physics question. Mathematically, all I can tell you is that if $f$ is a quadratic function (an inverted parabola), then $f''$ is constant, therefore it takes the same value at $10$ as it does everywhere else.
The answer given several years before this one completely answers the mathematical question. I will give some thoughts here on how this relates to the physical question.

The short answer is:

It is common in textbooks to gloss over what happens at the endpoints of actions like the thrown rock and to plot the graphs as if neither the thrower nor the ground exists.
The rock is just magically already rising at vertical velocity $v'(0)$ at time $t=0$ and we just stop looking at it when $t=10.$
We never actually model the rock hitting the ground, so we have nothing to say about what the acceleration might be when that happens.

---

If the displacement graph actually plots the rock as stationary before $t=0$ and after $t=10,$ that is, if it is a plot of the function

$$
y(t) = 
\begin{cases}
0 & t < 0, \\
-32t^2 + 320 t & 0 \leq t \leq 10, \\
0 & t > 10, \\
\end{cases} \tag1
$$

then the mathematical answer is that the velocity and acceleration are both *undefined* at $t=0$ and at $t = 10.$
The velocity and acceleration in this model are the first and second derivatives of the function in Equation $(1),$

$$
y'(t) = 
\begin{cases}
0 & t < 0, \\
160 - 32t & 0 < t < 10, \\
0 & t > 10, \\
\end{cases}
$$
$$
y''(t) = 
\begin{cases}
0 & t < 0, \\
-32 & 0 < t < 10, \\
0 & t > 10, \\
\end{cases}
$$

Both of these functions have "holes" at $t=0$ and at $t=10.$
The holes are the result of an incomplete physical model:
the function in Equation $(1)$ ignores the thrower's motion that accelerates the rock from zero velocity to vertical velocity $160$ upward,
and it ignores the interaction of the rock with the ground that decelerates the rock (usually in a very short period of time) from vertical velocity $-160$ to zero.

It is much simpler not to think about details like this, so in simple textbook exercises we don't.