If we're talking about ONLY the horizontal position against time, we do get a function that is nicely approximated by a quadratic.
I get x(t) = -0.4373t^2 + 3.9568t + 0.0490
The derivative of this is x'(t) = -0.8746t + 3.9568
This gives the first three pieces of velocity
x'(0) = 3.9568
x'(1) = 3.0822
x'(2) = 2.2076
On the other hand, the first three first differences of the given data suggest:
(0.8-0)/0.2 = 4
(1.56-0.8)/0.2 = 3.8
(2.29-1.56)/0.2 = 3.65
There is some discrepancy there. If you can come to terms with exactly when the two versions are measured, perhaps you can reconcile them.
Hint: Derivative are instantaneous and differences are averages.