Infinity:The headstart paradox

Re: Infinity:The headstart paradox

<div class='quotetop'>QUOTE(\"Alpha and 0mega\")</div>
Alrite,say,man A and man B wanted to have a race.So man A give man B a headstart cos man A is twice as fast as man B.When man B ran 10m,man A started running,the distance between them was 10m,after which the distance between them was reduce to 5m becos man A was twice as fast.then the distance was reduce to 2.5m,125cm,62.5cm,31.25cm and so on.if u were to keep on dividing the distance between them,its not gonna end,its infinite.So its seems that man A will never overtake man B.[/b]
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We'll say the head start was 2 seconds.

We'll say man A runs at 10 m/sec and Man B runs at 5m/sec

First second..... (the headstart)
Man A stands still and goes 0 m. He's gone a total of 0 m by the end of the second.
Man B goes 5 m. He has gone a total of 5m by the end.

Second Second (heheheh, get it? bad pun) (headstart continued)
Man A stands still and goes 0 m. He's gone a total of 0 m by the end of the second.
Man B goes 5 m. He has gone a total of 10m by the end.

Third Second.
Man A goes 10 m. He's gone a total of 10 m by the end of the second.
Man B goes 5 m. He's gone a total of 15 m by the end of the second.

Fourth Second
Man A goes 10 m. He's gone a total of 20 m by the end of the second.
Man B goes 5 m. He's gone a total of 20 m by the end of the second.

This is the catching point.

Fifth Second
Man A goes 10 m. He's gone a total of 30 m by the end of the second.
Man B goes 5 m. He's gone a total of 25 m by the end of the second.



The thing is, Alpha & Omega, the situation you have described is a contradiction because it could never result from one being twice as fast as the other.

That would require something much more complicated to happen, which I may describe in a later post, after I get back from the store (because it will soon close)
 
Re: Infinity:The headstart paradox

we'll say man B runs at a constant 10 m sec now.

Man A waits till man be has gone 30 m (so he waits three seconds)

Second #4
ManA takes off at 20 m second, for a total of 20 m traveled.
ManB continues at 10m second, for a total of 40 m traveled.

Second # 5 and #6
Man A drops his speed to 12.5 m sec, so by the end of second #6 he's gone 45 m
Man B continues at 10 m sec, so by the end of second #6 he's gone 60 m

Second # 7 and # 8
Man A continues at 12.5 m sec, so by the end of second #8 he's gone 70 m.
Man B continues at 10 m sec, so by the end of second #8 he's gone 80 m.

Second #9 and #10
Man A drops his speed to 11.875 m sec, so by the end of second #10 he's gone 93.75 m
Man B continues at 10 m sec, so by the end of second #10 he's gone 100 m.

Second #11 and # 12
Man A drops his speed further to 11.25 m sec, so by the end of second #12 he's gone 116.25 m.
Man B continues at 10 m sec, so by the end of second #12 he's gone 120 m.


See where this is going (I actually needed a calculator for that last one)? If the distance between the runners is really to diminish by half every round of running, Man A's speed must steadily be approaching that of Man B, getting slower and slower until they're running almost-but-not-quite the same speed.

If such a condition could be reached, then man A would never pass man B, though he would get close enough to tap him on the shoulder, and the race would be a real precision photo finish, of course.

But.... that's quite a condition.
 

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