Hello Ramona! I figured we could write this as an ordinary differential equation. If x(t) is the position of the ant at time t, and l(t) = 100 + 2 * t the length of the rubber, we have dx/dt = -1 + 2 * x(t) / l(t) Asking Wolfram Alpha to solve this for me (because I haven't done this type of exercise since ages): https://www.wolframalpha.com/input?i=dx%2Fdt+%3D+-1+%2B+2+*+x+%2F+%28100+%2B+2+*+t%29+with+x%280%29%3D100 The answer is x = t => (t + 50) * (-log(t + 50) + 2 + log(50)) Which means that the ant reaches the wall after 50(e^2-1) seconds (319.45s): https://www.wolframalpha.com/input?i=%28t+%2B+50%29+*+%28-log%28t+%2B+50%29+%2B+2+%2B+log%2850%29%29+%3D+0 We can also investigate when the ant starts going forward again, i.e. when the relative speed dx/dt above is equal to 0: https://www.wolframalpha.com/input?i=d+%28%28t+%2B+50%29+*+%28-+log%28t+%2B+50%29+%2B+2+%2B+log%2850%29%29%29+%2F+dt+%3D+0 and that's at time t = 50(e-1) ~= 85.914s; after that excruciating minute and a half of going backwards, our dear ant finds hope again and starts walking forward.

Oh wow!!! This is great — that is the right solution (from the comments in the tiktok video) but I didn't know quite how to derive it. Amazing!