Sunday
April 26, 2015

Homework Help: Physics

Posted by Nathan on Monday, August 15, 2005 at 8:42pm.

One end of a rubber band with a length L is attached to a wall. The rubber band is held horizontally and the loose end is pulled away from the wall at a speed v, stretching the band. Simultaneously, an ant begins to crawl away from the wall along the rubber band with a speed u < v. Assume that the rubber band can be infinitely stretched. Will the ant ever make it to the loose end of the rubber band? If so, how long will it take?


The speed of the ant is u+ speed of band. The speed of the band depends on where the ant is, at the end it is v, halfway it is v/2, and at position d from the wall it is d*v/(L+v*time)

So to get to the end, the ant has to go faster than the band below him. He is, because u+ d*v/(v*time)>velocity of band below him.

consider The position of the ant at time T.
Position= average velocity*Time
= [u + 1/2 (position*v/(v*T)*]T
=uT + 1/2 *position*vT/(VT)

solving this for position...
Position = uT/[1-vt/2(vT)]=2 uT
So, can position of the ant ever be greater than vT , the end of the band...
vT =? 2uT so it can be true if

u = v/2 or greater.

Please check my thinking


I have an error I forgot to delete... Here is the new..
The speed of the ant is u+ speed of band. The speed of the band depends on where the ant is, at the end it is v, halfway it is v/2, and at position d from the wall it is d*v/(L+v*time). This assumes constant velocity v for the end of the rubber band.(Acceleration is zero)

So to get to the end, the ant has to go faster than the band below him. He is, because u+ d*v/(v*time)>velocity of band below him.

consider The position of the ant at time T.
Position= average velocity*Time
= [u + 1/2 (position*v/(v*T)*]T
=uT + 1/2 *position*vT/(VT)

solving this for position...
Position = uT/[1-vt/2(vT)]=2 uT
So, can position of the ant ever be greater than vT , the end of the band...
vT =? 2uT so it can be true if

u = v/2 or greater.

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