Sure, but I don't think it'll be that much more accurate because the fit curve can deviate a lot.
Anyway, the first step is to find the equation of a curve through multiple points.
2 points would be a line,
3 would be quadratic
4 would be cubic
So basically you'd keep track of the last 4 positions and then using these cubic equations:
x(t)=a*t^3+b*t^2+c*t+d
y(y)=e*t^3+f*t^2+g*t+h
you'd plug in all the x,y and t values and solve for the unknowns. you can probably pick evenly spaced values for t. Here would be all your equations:
x0=d
y0=h
x1=a+b+c+d
y1=e+f+g+h
x2=a*2^3+b*2^2+c*2+d
y2=e*2^3+f*2^2+g*2+h
x3=a*3^3+b*3^2+c*3+d
y3=e*3^3+f*3^2+g*3+h
Then you'd just solve for all the unknowns: a,b,c,d,e,f,g,h which would define the shape of the cubic curve that goes through the last four points.
The solving could be done with algebra as a system of equations. you could also use a matrix and guass-jordon to solve it but the result would be the same. You can get someone else more current with math to do it if it's not something you want to do.
Anyways after that you'd have the cubic equations through the last four points, and going from t=0.0 to t=1.0 would give a curve from the current position to the previous. Actually that's flawed because if you drew that curve every frame the resulting curve would break every frame. so a better solution would be to use a different curve such as catmoll-rom or hermite or bezier. basically using p(0)-p(1) and p(2)-p(3) as tangents a only looking at the curve between 1 and 2. I've seen equations that do that and keep the curve continuous, but it probably can be derived.
So the above gets you the equations for a curve, or at least gives some leads how to get it. Next the actual getting of the length. To get it perfectly you could use an integral, but that's not solvable in most cases. The simplest solution is to use the equations we found to get a bunch of positions between frames and add the distances between those together. The more you use the more accuracy you'd get.