This should probably be on the math stackexchange. It's a differential equations problem.
The interesting part about the elastic band is that it provides the ant with a fraction of the car's velocity. The fraction varies depending on how far along the elastic the ant has gotten. For instance, at t = 0 when the ant is at the very beginning, the elastic is anchored and doesn't move. But at t = ? when the ant is halfway to the car, the elastic point is moving at half the speed of the car. And at t = ? when the ant catches up, the elastic is moving at the full speed of the car.
The distance of the car from the wall at time t is $d_c = \frac{1}{2} c t^2 + 1$
Let the distance of the ant from the wall at time t be $d_a$, which is a function of t
The velocity of the ant at time t is $d_a'$ (first derivative of $d_a$)
We know that $d_a' = a + \frac{d_a} {d_c} d_c'$ (the constant component a, plus a fraction of the car's velocity based on how far along the elastic the ant is at the time)
Substituting:
$d_a' = a + \frac{d_a} {(\frac{1}{2} c t^2 + 1)} (c t)$
That's a first order differential equation. I plugged it into Wolfram Alpha's differential equation solver. It provided a complicated solution:
$y = \frac{a(c t^2 + 2)\arctan(x\sqrt{\frac{c}{2}})}{\sqrt{2c}} + k_1(c t^2+2)$
$k_1$ is a constant that we have to provide based on the initial conditions. Our initial condition is y(0) = 0. Substituting that in, we see $k_1$ must be 0.
So we're left with $y = \frac{a(c t^2 + 2)\arctan(t\sqrt{\frac{c}{2}})}{\sqrt{2c}}$
The ant catches up to the car when $d_a = d_c$, or
$\frac{a(c t^2 + 2)\arctan(t\sqrt{\frac{c}{2}})}{\sqrt{2c}} = \frac{1}{2} c t^2 + 1$
That's too hairy to keep working with. But there are definitely solutions. Pick an x (time), a c (car's acceleration), substitute, and solve for a, the required minimum speed of the ant. For instance, t = 10, c = 1, results in a = ~0.494.
Hopefully someone can check my math, because I'm not sure I did all the math markup correctly!
A car starts distance 1 from a wall...
Does the ant's step cover more or less distance than (1)? $\endgroup$