One easy answer would be that in practical purposes, it is a very special sort of parameter. This point comes up quite clearly when we do the derivation for the work-energy theorem:
$$ W = \int\mathbf{ F} \cdot \text d\mathbf s = \int m \left(\frac{\text dv}{\text dx}\cdot \frac{\text dx}{\text dt} \right)\, \text dx = \int m v\, \text dv =\frac12 mv^2 +C$$
When we go from result of first equality to the one of second equality, we are implicitly choosing a very specific parameterization which is of the actual time experienced ticking as the object moves through the path, but, on a mathematical level, no matter how fast we would have run through the force field (assuming force is indep of time), it would be that the work integral gives same answer.
So, this made me wonder, what properties does time have beyond just being a parameter when doing integrals in Newtonian mechanics?