![]() |
![]() |
#1 |
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
23·661 Posts |
![]()
An ant is walking along a rubber band at 1 cm/Sec.
The band is stretched 2 cm/Sec. Does the ant ever get to the end? If so, how long does it take? What if the band is stretched 1000 cm/Sec? Clarification: The rubber band is straight not a round band. Starting length is 100 cm Ant starts at the beginning Last fiddled with by petrw1 on 2022-01-31 at 21:38 Reason: clarification |
![]() |
![]() |
![]() |
#2 |
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2×3×52×73 Posts |
![]()
Clarification question: Does the stretching measurement measure the total length of the band (y in the illustration) , or the length in a single axis (x in the illustration)?
|
![]() |
![]() |
![]() |
#3 |
If I May
"Chris Halsall"
Sep 2002
Barbados
67·167 Posts |
![]() |
![]() |
![]() |
![]() |
#4 | |
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
14A816 Posts |
![]() Quote:
Last fiddled with by petrw1 on 2022-01-31 at 21:37 |
|
![]() |
![]() |
![]() |
#5 |
If I May
"Chris Halsall"
Sep 2002
Barbados
67·167 Posts |
![]()
So, then, if I understand the puzzle correctly, this is basically a two-dimensional (x,t) version of the four D (x,y,z,t) "light cone" problem in an expanding Universe.
And, so... No. The ant will never reach the other end. I would be happy to be corrected if I am incorrect in this thinking. Last fiddled with by chalsall on 2022-01-31 at 21:50 Reason: Linewrap correction. |
![]() |
![]() |
![]() |
#6 |
Jul 2003
So Cal
2,621 Posts |
![]()
The ant's speed of 1 cm/s is measured relative to the rubber band? If so, then yes. The ant will eventually reach the end.
My approach is to find the ant's position and speed in the lab frame using a Galilean transformation, which relates the ant's speed in the lab frame to its position. This results in a first order differential equation, which gives the ant's position. Then use that to solve for the time at which the ant's position matches the length of the rubber band. This gives a finite but exponentially increasing time as the ratio of the speed of the stretching of the rubber band to the speed of the ant increases. With large ratios, this results in practical issues involving the lifetime of the ant, the elasticity of the rubber, the length of the rubber band, the the force required to continue stretching it. t = L/v (e^(v/a) -1) where L is the initial length of the rubber band, v is the speed of the stretch of the rubber band, and a is the speed of the ant relative to the rubber band. Quick calculation here. And apparently SPOILER and TEX tags don't mix! Last fiddled with by frmky on 2022-01-31 at 23:11 Reason: Add link |
![]() |
![]() |
![]() |
#7 |
Apr 2020
22·3·79 Posts |
![]()
It can be seen without calculus that the ant must reach the end. This approach also gives more intuition for why this happens.
Consider the discrete version of the problem, where the 2cm stretching is done all at once, on the second, every second, including at time 0. The ant travels strictly further in the continuous version than in the discrete version, because the further the ant is along the band, the more it is pulled forward by the stretching, so if all the stretching is done at the beginning of each 1-second interval when it is slightly further back, it will not be pulled quite as far. Now consider the proportion of the band that the ant has covered after t seconds: it is 1/102 + 1/104 + ... + 1/(100+2t), which diverges as t goes to infinity, by the standard proof of divergence of the harmonic series. So it certainly reaches 1, i.e. the ant gets to the end. Similarly, for initial length L and stretching speed v, we get 1/(L+v) + 1/(L+2v) + ... + 1/(L+tv), which once again diverges. As the ant travels slightly faster in the continuous version, it must reach the end in this case too. |
![]() |
![]() |
![]() |
#8 | |
If I May
"Chris Halsall"
Sep 2002
Barbados
67·167 Posts |
![]() Quote:
I'm a programmer by training. I rely on those trained in maths for that domain. I greatly appreciate the continuous training we all receive here for those interested by simply observing. And those willing to ask (and learn from) serious questions and their answers. In many different domains of discipline. |
|
![]() |
![]() |
![]() |
#9 |
Feb 2017
Nowhere
2·3,167 Posts |
![]()
I got the same formula as frmky by a different approach.
I considered the fraction f of the rubber band's length between the ant and the starting point. We have f = 0 at t = 0. The ant reaches the end when f = 1. I noticed that if the ant were not moving, f would remain constant. This led me to consider how fast f would change when the ant moved, and - boom! EDIT: After a bit more thought, I realized that I was implicitly assuming that the rate of stretching was uniform along the entire length of the rubber band. This seems the only reasonable default assumption in the absence of any explicitly stated alternative. Last fiddled with by Dr Sardonicus on 2022-02-01 at 00:33 Reason: As indicated |
![]() |
![]() |
![]() |
#10 |
If I May
"Chris Halsall"
Sep 2002
Barbados
67·167 Posts |
![]()
This is why I listen to experts.
Empirically, my intuition is often incorrect. And... Many (many!) others know more than I do... If I may please go completely orthogonal, Lennox entered my mind this evening. |
![]() |
![]() |
![]() |
#11 | |
"Oliver"
Sep 2017
Porta Westfalica, DE
5A516 Posts |
![]() Quote:
![]() |
|
![]() |
![]() |
![]() |
Thread Tools | |
![]() |
||||
Thread | Thread Starter | Forum | Replies | Last Post |
Rubber band puzzle. | Uncwilly | Puzzles | 17 | 2014-03-07 00:31 |