Okay, I'll pick up where Nuxy left off.
V (= dx/dt) = Vr + C(x/(Ct+L)) = Vr + x/(t+L/C)
Writing this out a little differently, we get: dx/dt - 1/(t+L/C)*x = Vr
Now, I'm going to define a cute little function as: M(t) = 1/(t+L/C)
I multiply our previous equation by my new function: M*dx/dt - (M/(t+L/C))*x = M*Vr
Which happens to be (feel free to check this, but I defined it this way on purpose): M*dx/dt + dM/dt*x = M*Vr
Using a well-known law which I don't know the English name for (product rule! thanks, nux!):
d(M*x)/dt = M*Vr = Vr/(t+L/C)
Integrating both sides (according to t, of course), gives: M*x = Vr*ln(t+L/C)
We then need only divide by M to get: x = Vr*(t+L/C)*ln(t+L/C)
In order to find out when the ant will reach the car, let's equate!
Vr*(t+L/C)*ln(t+L/C) = Ct + L = C*(t+L/C)
Vr*ln(t+L/C) = C
ln(t+L/C) = C/Vr
t + L/C = e^(C/Vr)
t = e^(C/Vr) - L/C
Substituting the given Vr = 1, C = 2, L = 2, we get: t = e^2 - 1 seconds
And for the second part (Vr = 0.5, C = 2, L = 2), we get: t = e^4 - 1 seconds
Yes - so long as the ant is moving, no matter how slow, it will eventually reach the car. Pretty neat. ^_^ So, am I right?
(By the way, I'm mildly disappointed in you, Nux. You said you love math! ^_~)