is the initial velocity just the average of all the velocities?
I mean here is a little of my data
Velocity Stopping Distance
1.18 .22
1.58 .32
1.27 .295
1.12 .151
1.33 .28
velocity squared
1.392
2.49
1.61
1.25
1.76

In this example, did you push the object 5 times? (This is what I've understood so far).
What does the velocity stand for? Is it the velocity you pushed the object? If yes then it's the initial velocity of the object and its velocity will decrease until the object stops.
I'll do as if it was true. First example :
$\displaystyle a(t)=a$.
$\displaystyle v(t)=a\cdot t+1.18$.
$\displaystyle x(t)=\frac{a\cdot t^2}{2}+1.18\cdot t$. But we know that when $\displaystyle x=0.22$, the velocity is worth $\displaystyle \frac{0m}{s}$.
So let's find at what t it occurs. The best way to do that is to count how much photos took the photogate from the instant you pushed the object till it stops completely. Say it had time to take 20 photos with time intervals of $\displaystyle 0.1 s$. This means the body stopped at $\displaystyle t=2s$.
So now we now that at $\displaystyle t=2s$, the velocity is worth $\displaystyle \frac{0m}{s}$. Thus $\displaystyle v(2)=2a+1.18=0 \Leftrightarrow a=\frac{1.18}{2}=\frac{0.59m}{s^2}$ in this case. Do that for all the values you got and you obtain all the accelerations it seems they are asking you.
But still, I don't understand why did you calculate the velocity squared, so I must be missing something. I'm almost sure I don't understand well what you've done.