The ant on a rubber rope paradox

Let the ant's position as a function of time be $x$ . At first it seems that the ant's movement is simply $\frac{dx}{dt}=\alpha$ , but this doesn't factor in the stretching of the rope. As the rope extends the section of rope behind the ant will extend as well, pushing the ant forward an additional amount.

As end of the rope stretches forward at a speed of $v$ , the ant will move forward at some fraction of $v$ equal to the fraction of the rope the ant has covered, since the stretching is distributed equally across the entire rope. At any point in time, the rope has a length of $c+vt$ , and so the ant will have gone a fraction of $\frac{x}{c+vt}$ . Therefore the speed of the ant due to the stretching is $\frac{vx}{c+vt}$ and so the total speed is $\frac{dx}{dt}=\frac{vx}{c+vt}+\alpha$ .

This results in a linear first-order differential equation:

$\frac{dx}{dt}-\frac{v}{c+vt} x=\alpha$

To solve this, we first find the integrating factor, $\mu(t)=e^{\int p(t)dt}$

$\begin{aligned}\mu(t)&=e^{\large\int -\frac{v}{c+vt} dt}\\ &=e^{-\ln(c+vt)+C_0}\\ &=\frac{e^{C_0}}{e^{\ln(c+vt)}}\\ &=\frac{K}{c+vt}&\text{Where the constant }K=e^{C_0}\end{aligned}$

Then we multiply the integrating factor through the differential equation, resulting in a product rule form on the left hand side:

$\begin{aligned}\frac{K}{c+vt}\frac{dx}{dt}-\frac{Kv}{(c+vt)^2} x&=\frac{K\alpha}{c+vt}\\ \frac{K}{c+vt}\cdot\frac{d}{dt}(x)-\frac{d}{dt}\left(\frac{K}{c+vt}\right)\cdot x&=\frac{K\alpha}{c+vt}\\ \frac{d}{dt}\left(\frac{K}{c+vt}\cdot x\right)&=\frac{K\alpha}{c+vt}\\ \frac{Kx}{c+vt}&=\int\frac{K\alpha}{c+vt}dt\\ &=\frac{K\alpha}{v}\ln(c+vt)+C\end{aligned}$

When $t=0,\,x=0$ , so

$\begin{aligned}\frac{K\alpha}{v}\ln(c)+C&=0\\ C&=-\frac{K\alpha}{v}\ln(c)\\ \therefore\frac{Kx}{c+vt}&=\frac{K\alpha}{v}\ln(c+vt)-\frac{K\alpha}{v}\ln(c)\\ \frac{Kx}{c+vt}&=\frac{K\alpha}{v}\ln\left(\frac{c+vt}{c}\right)\\ x&=\frac{\alpha}{v}(c+vt)\ln\left(1+\frac{vt}{c}\right)\end{aligned}$

Now we need to find when the position of the ant equals the position of the end of the rope, so:

$\begin{aligned}\frac{\alpha}{v}(c+vt)\ln\left(1+\frac{vt}{c}\right)&=c+vt\\ \ln\left(1+\frac{vt}{c}\right)&=\frac{v}{\alpha}\\ 1+\frac{vt}{c}&=e^{v/\alpha}\\ \frac{vt}{c}&=e^{v/\alpha}-1\\ t&=\frac{c}{v}\left(e^{v/\alpha}-1\right)\end{aligned}$

Therefore $A=1,B=1,C=-1$ , so the answer is $(1+1)^{-1}=\boxed{0.5}$