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Original Problem:

An ant starts from a rope of length 1m and with a speed of 0.1m/sec and simultaneous rope expands at a rate 1m/sec. If we find the ratio of end points to ant's position we get a sequence that converges to 1 (check it for yourself) and hence this implies that ant reaches the end point at some large enough time t.

Variation: Instead of starting at 0, the ant starts at 0.1m. Now the ratio of end point to ants position is fixed ie 1/10 and remains same throughout and hence never reaches 1, therefore the ant never reaches the end point. So here is the counter-intution, why come starting ahead be a disadvantage?

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    $\begingroup$ if the rope expands uniformly then the ratio is not fixed (the ant will still reach the end). am I missing something ? $\endgroup$ – Fabich Oct 22 '16 at 23:00
  • $\begingroup$ Yes, its not fixed, I am considering it for time intervals of 1 sec, ie ration after 1 sec, then 2 sec and so on which will give me a series which converges to 1. $\endgroup$ – Manish Kumar Singh Oct 22 '16 at 23:06
  • $\begingroup$ But the whole reason the other problem works is because the ant moves forward, increasing the ratio every time $\endgroup$ – Areeb Oct 22 '16 at 23:29
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    $\begingroup$ "and with a 0.1m/sec" - so not clear $\endgroup$ – paparazzo Oct 23 '16 at 0:36
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    $\begingroup$ How could the ratio be fixed? That corresponds to the case where the ant doesn't move at all, just rides along as the rope stretches. The ant, as it is moving, is going faster than an ant that is just riding along. Indeed, this case works out just like the original one. $\endgroup$ – Milo Brandt Oct 23 '16 at 1:52
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Your question currently doesn't contain enough information to understand what you tried, so I'll just solve the original puzzle and hope that this helps clear the confusion:

Let the relative position of the ant be $p = \frac{\text{position of the ant}}{\text{length of the rope}}$.

If the ant stays still, then it will remain at the same relative position in the rope. But because it's actually moving with a speed of 0.1 m/s at any given time, in a given moment of time $t$ the relative position must change with a speed of $\frac{\Delta p}{\Delta t} = \frac{0.1}{\text{size of the rope}} = \frac{0.1}{1+t}$.

Because I want to keep this to high-school math, let's round the speed down to $\frac{0.1}{2}$ in the first second, $\frac{0.1}{3}$ in the second, and so on. Because we rounded the speed down, we can say that the ant walks at least $\frac{0.1}{2} + \frac{0.1}{3} + ...$

This is related to the harmonic series, and unlike what you claimed in the question it does not converge to 1, rather it diverges to infinity. So given enough time, the ant will overtake the rope no matter how far from the end it was initially.

As you can see, I didn't even use the starting position of the ant in the solution, because it really doesn't make a difference to the answer.

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If I got your question right...

Ant will never reach the end in the first scenario itself

Here's why...

Let

t => The clock time
A => Position of ant at time t
L => Length of the rope at time t
D => (Remaining) Distance ant has to travel

D = (L-A) meters

T => Time it will take ant to reach end of the rope at time t

T = (Remaining) Distance ant has to travel / velocity of ant (ie 0.1m/sec)
T = D/0.1

At,
t=0 => A=0.0, L=1, D=(1-0.0), T = 10 sec
t=1 => A=0.1, L=2, D=(2-0.1), T = 19 sec
t=2 => A=0.2, L=3, D=(3-0.2), T = 28 sec
t=3 => A=0.3, L=4, D=(4-0.3), T = 37 sec
t=4 => A=0.4, L=5, D=(5-0.4), T = 46 sec

I think this is what needs clarification

Ratio does not reflect the distance (L-A) ant has to travel nor the time (D/0.1) it will take to cover that distance.

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  • $\begingroup$ if you read your answer carefully the difference between successive T reduces which gives the hint that it might converge at some later time. Alternately think in terms of percentage and tabulate your recordings, you will notice it slowly approaches 100%(this is what i did). $\endgroup$ – Manish Kumar Singh Oct 23 '16 at 3:55
  • $\begingroup$ No it doesn't. I checked. With every passing sec, a TDiff of 9 secs keeps adding up to the time it will take ant to reach the destination implying the ant keeps getting farther and farther away from the end point and will never reach it $\endgroup$ – WeShall Oct 23 '16 at 5:23
  • $\begingroup$ see answer in wiki $\endgroup$ – paparazzo Oct 23 '16 at 10:55

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