The number 1 rule when analysing experimental data is:
- Draw a graph
And the number 2 rule is
- If possible draw a straight line graph
Let's obey rule 1 and graph your data. We get:

OK, it looks like a quadratic as expected. But because it's not a straight line graph it's hard to see whether it's a quadratic or not. To get a straight line graph we have to play around with the data. We expect the distance:time equation to be:
$$ s = ut + \tfrac{1}{2}at^2 $$
Suppose we assume $u=0$, then this simplifies to:
$$ s = \tfrac{1}{2}at^2 $$
So if we graph $s$ against $t^2$ we should get a straight line with a gradient of $\tfrac{1}{2}a$. Well, let's do this. The result is:

The red dots are your data, and the blue line is a straight line that I've drawn to give what I think is the best fist to the points. And, well, the blue straight line looks a pretty good fit. The gradient of the line is $0.158$ m/sec$^2$, and this is $\tfrac{1}{2}a$ so we get:
$$ a = 0.316 \,\text{m/sec}^2 $$
But is this really the best fit? Is $u$ really zero? Well we can take our equation:
$$ s = ut + \tfrac{1}{2}at^2 $$
An divide through by $t$ to get:
$$ \frac{s}{t} = u + \tfrac{1}{2}at $$
This tells us that if we graph $s/t$ against $t$ we should get a straight line with a gradient of $\tfrac{1}{2}$ and a $y$ intercept of $u$. OK, let's try it. We get only three points this time because the first point gives $s/t = 0/0$, which we can't do. Anyhow, with three points the graph looks like:

Again the red dots are your data and the blue straight line is my fit. Although it's a long extrapolation back to the $y$ axis it looks to me as if $u$ is definitely not zero. In fact when I measure the gradient and intercept of my straight line I get:
$$\begin{align}
a &= 0.291 \,\text{m/sec}^2 \\
u &= 0.30 \,\text{m/sec}
\end{align}$$
Finally, lets go back to your original graph and show your data along with the curve we get if we use the values of $a$ and $u$ above. The graph now looks like:|

I'd say that was a pretty good fit!