Generalized "Worm on the rubber band " problem I found this « Worm on the rubber band » problem in Concrete Mathematics book.
A slow worm $W$ starts at one end of a meter-long rubber band and crawls one centimetre per minute toward the other end.
At the end of each minute, a keeper of the band $K$ stretches it one meter.
Does the worm ever reach the finish?
The given solution is: $W$ reaches the finish if and when $H(n)$ ever surpasses 100, where $H(n)$ is the $n$th Harmonic number.
How to solve this generalized problem with continuous data and with $W$ crawling with a velocity $u=f(t)$ and $K$ crawling with velocity $v=g(t)$  where $u(t)$ and $v(t)$ are both arbitrary functions of time.
For example, I want to find if and when the worm will reach the end of a rubber band of length $L$ if (with $a$ and $b$ constants) $u(t)=a*t$ and $v(t)=b*(t)$ ?    
 A: Bonjour, Jean-Pierre, the relative position of the insect on the rubber band is 
$$r(0)=x_W(0) / x_K(0)$$ 
and you will have to know it to solve the problem. It will be a number between 0 and 1.
Now, you want to find $t$ such that $x_W(t)=x_K(t)$ i.e. $r(t)=1$. To find it, you must calculate the time derivative of $r(t)$.
$$r'(t) = \frac{u(t)}{x_K(t)} . $$
It's because the relative position of the insect is determined essentially by its velocity $u(t)$ but the longer the rubber band is, i.e. the greater is $x_K(t)$, the less the insect manages to change the relative position.
Given your data, $x_K(t)$ is the indefinite integral (that you have to calculate) of the function $v(t)$ you have mentioned and $u(t)$ is something you mentioned. So both the numerator and denominator are, in principle, known functions of $t$, so you solve the differential equation above.
One may also write the general compact formula for a given $r(0),u(t),v(t)$. But it's more interesting to know a major example. If $v(t)$ is constant, $x_K(t)$ goes like $Kt+x_0$ and if $u(t)$ is constant, the equation above says $r'(t)\sim 1/(t+t_0)$. This is solved by $r(t)\sim \ln((t+t_0)/T)$ and when you invert it, you will find out that the $t$ for which $r(t)=1$ is an exponential function of the parameters. It's a straightforward exercise, and given the risk that this is a homework, I don't want to write every detail and the final answer here.
So the ant will finally get there but you will need an exponentially long time! For more general functions, the computation is more complicated.
A: Thanks to Lubos, I get the following results:
With the data of the question:  rubber band of length L, u(t)=a*t and v(t)=b*(t)  with a et b constants, the insect will ALWAYS reach the end of a rubber band in a time:
T= [(2.L)/b * ( Exp(b/a) -1) ] ^(1/2)
If u is constant and v= b(t) with b constant, the insect will reach the end in a time
T= (2.L/b)^(1/2)  *  tan[ ( b.L/2.u²)^1/2] 
ONLY  if  b < (Pi.u )² / 2.L
