Will the Bug Ever Reach the End of the Stretching Rubber Band?

AI Thread Summary
The discussion revolves around a mathematical problem involving a bug crawling on an infinitely stretching rubber band. The bug starts at the wall and moves towards the other end at 1 cm/sec, while the rubber band is extended by 1 km every second. Participants explore the differential equation governing the bug's position and suggest methods for solving it, including integration techniques. There is some confusion regarding the interpretation of the stretching of the rubber band, whether it is continuous or instantaneous. Ultimately, the consensus is that the problem can be approached through the ratio of the bug's position to the total length of the rubber band, leading to a solvable first-order ordinary differential equation.
  • #51
Suppose if the bug would have started walking from the end of the rubber band which is in our hand to end which is attached to wall? Will it reach the other end of the rubber band?
Assuming that other conditions of question remains same.
 
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  • #52
Satvik Pandey said:
Suppose if the bug would have started walking from the end of the rubber band which is in our hand to end which is attached to wall? Will it reach the other end of the rubber band?
Assuming that other conditions of question remains same.

What are your thoughts?
 
  • #53
Orodruin said:
What are your thoughts?

I think that bug will reach the other end of the band.
I think relative velocity of bug w.r.t Earth at any point ##x## (measured from the wall)
is ##u-\frac { xv }{ L+vt }##.
 
  • #54
Nathanael said:
\frac{dy}{dt}=\frac{\dot x}{L+vt}-\frac{xv}{(L+vt)^2}

(With initial value y=0 when t=0)
Which could be written as:

\frac{dy}{dt}=\frac{\dot x-yv}{L+vt}

(I got stuck here for a few minutes but then thought of this :smile:)
Maybe I should plug in \dot x=u+vy and then I get:

\frac{dy}{dt}=\frac{u}{L+vt}
Excellent! This is what I wanted you to do way back in post #7. It's such a simple result that it's hard to give a hint (other than try it, it's easy!) that doesn't tell the answer.
 
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