Nonuniform acceleration due to rubber rope What I want:
I have a rubber rope which is $5m$ in length when not stressed and is able to stretch about $100\%$ (to $10m$ long). I want to accelerate a constant mass horizontally, which has negligible friction. I'd like to have a function that tells me the velocity of the mass dependent on time, so for instance velocity $1 s$ after releasing it.
What I did:​
I've done some measurements of forces of the rope when pulling it to different lengths. Of course, when pulling $0cm$ (total length $5m$) I got a force of $0N$. Here is a graph of my results.

$x-axis$: displacement of one end of the rope
$y-axis$: measured force
I was also able to do a regression and found a function which describes how much force I get after I pull a given length.
I name this function $F(s)$ for Force dependent on displacement.
From this, it's easy to get the acceleration function, which is $a(s) = F(s)/m$ with $m = mass$ of the object I want to accelerate.
But now I'm stuck.
I somehow need to get $a(t)$ instead of $a(s)$, thus the acceleration by time, not by length, so I can then integrate that to get $v(t)$.
How do I convert the dependency of the function?
 A: You certainly want to numerically integrate your motion equation givent your expression of the force.
I will assume that the mass $m$ is attached at one side of the rope, while the other side is attached to a wall or something that won't move during the integration. Something like this, where the spring is in fact your rope, with $x$ the extension of the rope

Then you can simply integrate (for example using Euler scheme) the equation $$\frac{d^2x}{dt^2} = \frac{F(x)}{m}$$ Here is some Ruby code I use whenever I want to get numerical solution of an ODE in mechanics:
##################################################
## Integration using Euler (mid-point) method knowing position x and speed xp=\dot{x}
##################################################

def integration(x,xp)
  dxp = force(x)*@dt
  dx  = (xp+dxp/2)*@dt
  return x+dx,xp+dxp
end


##################################################
## Force expression knowing position x
##################################################

def force(x)
  return -10*x
end


##################################################
## Effective integration loop
##################################################

@dt = 0.001  # time increment
t = 0  # initial time
x = 5  # initial position
xp= 0  # initial speed
10000.times do |i|
  t += @dt
  x,xp = integration(x,xp)
  puts "#{t}\t#{x}\t#{xp}"
end

Just plug in your initial conditions and force expression from your measures and you will have both position and speed along time (NB: I took $m=1$ in my code but you can add the correct value wherever you like in the integration function or in the force function [that shall be renamed "acceleration" then]).
You can also refine it to integrate over constant total time and output fewer results than you have integration steps (precision increase with smaller integration step @dt but it can become difficult to draw when you have a million points just to be more precise in the integration)
A: When the acceleration is a function of position use the following
$$ a(x) = \frac{{\rm d}v}{{\rm d}t} = \frac{{\rm d}v}{{\rm d}x} \frac{{\rm d}x}{{\rm d}t} = \frac{{\rm d}v}{{\rm d}x} u $$
$$ \int a(x)\,{\rm d} x = \int u\,{\rm d} u = \frac{1}{2} u^2 + K_1 $$
which is solved for $u(x)$.
The the position is found from
$$ t = \int \frac{1}{u(x)}\,{\rm d} x + K_2 $$
which is solved for $x(t)$.
Example
$$ a(x) = A\,x^{b} $$
$$ \int A\,x^{b}\,{\rm d} x = \frac{A}{b+1} \left( x^{b+1}-1\right) = \frac{1}{2} u^2 + K_1 $$
when $u=0$ at $x=0$ then $K_1 = \mbox{-} \frac{A}{b+1}$ or
$$ u(x) = \sqrt{\frac{2 A}{b+1} x^{b+1} } $$
Then 
$$ t = \int \frac{1}{\sqrt{\frac{2 A}{b+1} x^{b+1} }}\,{\rm d} x + K_2 = \sqrt{ \frac{2(b+1)}{A (b-1)^2}} \left( x^{\mbox{-}\frac{b-1}{2}}-1\right) + K_2  $$
and when $x=0$, $t=0$ then $K_2 = \sqrt{ \frac{2(b+1)}{A (b-1)^2}}$ or
$$ t = \sqrt{ \frac{2(b+1)}{A (b-1)^2}} x^{\mbox{-}\frac{b-1}{2}} $$
$$ x(t) = \left( \frac{t}{\sqrt{ \frac{2(b+1)}{A (b-1)^2}}}\right) ^{\mbox{-} \frac{2}{b-1}}  = \left( \frac{A (b-1)^2}{2 (b+1)}\right)^{\mbox{-}\frac{1}{b-1}} \;t^\left({\mbox{-}\frac{2}{b-1}}\right)$$
If your mass was $m=1$ then $a=f(x)$ in graph and
$$ x=1.397882\, t^{3.058906} $$
