If I apply a constant force to an object until I've reversed its starting velocity, does its final position remain unchanged? 
A body of mass $m$ has initial velocity $v_0$ in the positive $x$-direction. It is acted on by a constant force $F$ for time $t$ until the velocity becomes zero; the force continues to act on the body until its velocity becomes $−v_0$ in the same amount of time. Write an expression for the total distance the body travels in terms of the variables indicated. 

I let $t'$ denote the time at which the velocity goes to zero so I can use $t$ as a variable. Then I found
$$a = \frac{\Delta v}{\Delta t} = \frac{-2v_0}{2t'} = -\frac{v_0}{t'}$$
Substituting this expression for $a$ into $s(t) = \frac{1}{2} a t^2 + v_0 t$ yields
$$\begin{align}
s(t) & = \left(-\frac{1}{2} \frac{v_0}{t'}\right) t^2 + v_0 t \\
s(2t') & = \left(-\frac{1}{2}\frac{v_0}{t'}\right)(2t')^2 + v_0 (2t')^2 = 0
\end{align}$$
This makes intuitive sense to me. If the positive $x$ direction is up and the force is gravity, then if we shoot a ball upward with velocity $v_0$, it will decelerate to velocity $-v_0$ exactly when it returns to the launch position.
To double check, I tried taking (from $a = -\frac{v_0}{t'}$ and $F = ma$) $a = \frac{F}{m}$ and $v_0 = -\frac{F}{m}t'$. I get the same answer when substituting into $s(t)$:
$$\begin{align}
s(t) & = \frac{1}{2} a t^2 + v_0 t \\
& = \left(\frac{1}{2}\frac{F}{m}\right) t^2  -\left(\frac{F}{m}t'\right) t \\
\implies s(2t') &= \left(\frac{1}{2}\frac{F}{m}\right) (2t')^2  -\left(\frac{F}{m}t'\right) (2t') =0
\end{align}$$
My textbook, however, says without comment that the answer is $\frac{F}{m}(t)^2$, or in my notation, $\frac{F}{m}(t')^2$. Am I wrong, or is the textbook wrong, or could this be an ambiguity in wording—e.g. is there a way to construe "total distance the body travels" as "the furthest position the body reaches from its starting point"? Or could it be that the question is asking not about the object's location at time $t = 2t'$ but about its location as a function of time $s(t)$? In that case my answer is found above and would still be wrong.
 A: The total distance is a sum of two distances: $S_{1}$ and $S_{2}$, where the first is till the velocity goes to zero, the second: from zero to $-V_{0}$. $S_{1}=V_{0}t+\frac{1}{2}at^{2}$. Here $a=\frac{-V_{0}}{t}$. So $S_{1}=V_{0}t-\frac{V_{0}t}{2}=\frac{V_{0}t}{2}$.
The same for $S_{2}$ but with zero initial speed and positive acceleration: $S_{2}=\frac{1}{2}at^{2}=\frac{1}{2}\frac{V_{0}}{t}t^{2}=\frac{V_{0}t}{2}$
Total $S=S_{1}+S_{2}=V_{0}t=\frac{F}{m}t*t=\frac{F}{m}t^{2}$.
A: The textbook's answer reflects the total distance traveled, i.e.
$$\vert s(t') - s(0) \vert + \vert s(2t') - s(t') \vert = \vert \frac{1}{2} v_0 t' - 0 \vert + \vert 0 - \frac{1}{2} v_0 t' \vert = v_0 t' = -\frac{\vec F}{m}(t')^2 = \frac{F}{m}(t')^2$$
I prefer this way of organizing it because it lets me treat the displacement as a single function $s(t)$ and preserves the distinction between a specific value $t'$ and the variable $t$ that I am accustomed to in mathematics. Others may find the answer from @Anton Baranikov more legible.
