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crook
Member •
Apr 16, 2007
To the best of my knowledge, the problem was discussed in Mindsport section of TOI! heck I don't know the answer.
I think the ant should take infinite time to reach the other end.
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Since the ant is located on the rubber, the total distance that the ant has to cover is just 100 cm. The total time taken will be 100 cm / 1 (cm/sec) = 100 sec.
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@the_big_K
Is the answer given by rocker is correct ???
This is what i think.
The aunt is at a fixed point P.Now the other end,Q is moving away from aunt(or point P) at the speed of 100 cm/sec.So in one second,the point Q will be 100cm that means Q is now 200cm apart from P.And after one more second,it will be at 300cm apart.Thus the distacne PQ will go on increasing.And the speed of aunt is not sufficient to compensate for that.
Am i right or am i misinterpreting the puzzle.
Anyone please clarify..
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This problem can be correlated to a person walking@1m/s on a 100m long train.the person will cover the length of the train in 100s irrespective of the speed of the train.
Now people will say that here the entire rubber band is not moving, it is getting stretched.Hence the ant which is moving at an absolute speed of 1cm/s will be assisted by the stretching of the rubber band (increasing its absolute speed) and will cover the distance in 100s.(the increasing of speed can be justified if you look at stretching on a microscopic level)
😁
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The ant never reaches the end Q.....
bcoz the band is infinitely elastic rubber band......
So it keep on stretches......😁
Do u think, It is correct??????
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The problem has not been interpreted correctly. Assuming uniform elasticity, the length of distance the ant has to cover is increasing by 100 cm/s ... but please note, that in the first second, the ant has covered 1cm, the band is 200cm long ... and the ant has covered 2cms (since that 1cm is now 2cm ... due to stretching!!).
Then the ant goes ahead by 1cm, the band becomes 300cm long .. the distance behind the ant is 5cms. Then it goes ahead by 1cm, the band becomes 400cm long ... the distance behind the ant is 8 cms ahead.
Time Distance Covered %age
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1 100 1 1%
2 200 3 1.5%
3 300 5 1.66%
4 400 8 2%
Looks like the ant is making some progress after all 😀. I say it will reach the end. And you know what ...
Moral of the story - Determination gets you there.
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banu
Member •
Aug 11, 2008
no it won't reach Q
bcoz it will fall down as the band is stretched.
banu
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Yeah i agree with the solution provided by Kidakaka...
Since the rubber is uniform ....so the distance travelled by the ant also get's streched.....
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distance -time relation for ant:
(1)t*(100/100)*t
as per second increment of unit section rubber length is (100/100)*x where x is its orignal length
therefore increment in given section in time t is:
1*x*t
hence length reletion with time is =100t
left distance=100t-t*t=let's say "r"
ant rech at Q when
100t-t*t=0
therefore
t=100sec
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Very nice answer Kida!!
Its beyond imagination!!!
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no it won't reach the Q
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ya it's beyond the imagination.
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wat could be the answer then
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Very very interesting question!! I strongly believe that the ant will reach the end at one stage as Kida explains, but it will take a very very long amount of time.. The fact we have to consider here is the ant's position relative to the stretching of the rubber band. I dont know how to explain it, but i'm pretty sure it will make it! I would love to see someone bring out any solid proof..
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The ant will surely reach the end at some point of time. As the band is being stretched, even the ant's position from the starting point increases at the rate of 100 cm/sec. Now in the same 1sec, the ant has covered a dist. of 1 cm. So actually the ant is covering a distance of 101 cm/sec. I don't know the exact time but it should reach.
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The ant will reach the end, albeit after a very long time. Will post a detailed mathematical solution once I complete it.
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Its simple.The ant would take same amount of time if the rubberband had not been streched.
Because,speed of ant would be (100+1)cm/sec.
For example:If we are walking inside a in a moving train our net velocity is algebraic addition of{velocity of train+our own velocity};-)
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