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Worm Problem



OK here's my take on the problem.
at time step t : worm at position w moves to w + v dt
( dt =1 for discrete case)
Now the rope expands from L to L + V dt ( note upper/ lower case V, v )
Since this is a uniform expansion, the worm position is now w' where
w' = (w + v dt) * {(L + V dt)/L }
Now compute the FRACTION of the whole distance traveled
f (at t+dt) = w' / (L + V dt) = f(at t) + v dt / L
also L = 1km + V t
so we get a differential equation df = (v/V) dt/(t + 1km/V)
Integrating ( df from 0 to 1), we get 1 = (v/V) ln(T)
or T = exp(V/v) = exp(10^5)

of course, this is the same result as has been posted since
exp(10^5) = 10^(log(e)*(10^5)) , and log(e)= 0.434294