1ya fine ..no problem with that :D
Let A is behind B in a race.B is having initial lead of 's' and velocities of A and B are VA and VB respectively.Given that VA > VB . We know sometime later A will overtake B. But , what is wrong with the following argument ?
When A reaches where B was initially , B would have moved further away.(VB ≠0).Again when A reaches the next position where B was , B would have moved further away.In this manner A would never catch up B.What's wrong ?
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28 Answers
1The problem is "concept of convergence" was not thr when this paradox was put up by greek mathematician Zeno.He could only see there will be infinite steps before A catches B , but he could not see that the time will converge and become finite...(infinite steps but finite time :D)
1yes..the question really seemed to be quite paradoxical..
Satya...i cudnt understand wat time convergence means...
though, i was getting an almost similar result myself..
First of all, we do actually get a perfect decreasing GP in both distance travelled and time taken...thus, sum in either case must be finite..
then, finding the sum of GP to infinite terms, we obtain the same result that we wud hav obtained using conventional kinematics...
1Ok i think i should probably explain now..
t1 = s/VA ..distance moved by B is VB(s/VA)
t2 = sVB/VA2 ...distance moved by B will be sVB2/VA2
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Σt = (s/VA)/(1 - VB/VA) = s/(VA - VB)
So we can see that the time converges..
The problem with the argument is that time is considered to be discreet here wheres as in reality it is continuous and A overtakes B in infinite steps but not infinite time.This paradox was proposed by Zeno(a greek mathematician) when the concept of convergence of infinite series was not there ... :D
9i think u r concerning urself with time before catching up
if u take limit of the time intervals , it becomes finite and not infinite
1relative vel btwn A and B : Va - Vb
relative distance : s
so time taken for A to overtake B: t= s/ (Va - Vb)
dist moved by A whr it overtakes B: Va . t
y should it be infinite time ?
there's a finite afetr which we can see A overtaking B.
i dun get you :(
post your soln..
1Do u guys need some more time or u want me to answer this ?? :D
1ya i think so it is right if we consider sepration b/w thm at any time
11(Va-Vb)t is xtra distance covered by A. So dist bet them at t seconds after separation bet them was s is what i've given
11Every time A reaches B's previous position, the gap between them will get smaller and smaller
1acc to me in the st the gap dist is nt considered nd compared with the initial gap it is the dist b/w A nd B tht is considered leaving aside the already gap tht ws thr
11More clear ans pls, honey [1]
not able to understand a word of what u r trying to say
1since 'B' is already ahead of 'A' at a dist of 's' so in your st u r considering initial position of B which is ofcourse ahead of A and whnever A reaches thr B hs already moved 4m thr nd this dist wld be thr till A reaches equal to B and it will happen as VA>VB
11the statement would not be valid for all points b reached
the time interval between a and b reaching the point would be getting smaller .
so ultimately depending upon Va-Vb A would catch up with B .
11If st is separation at time t, then
st=s-(vA-vB)t
Correct?[7]
1Dude there is a perfect GP series for the time taken by A in each step. I will give you guys some time to figure out this . :D
11Distance covered in each step is the same and is not decreasing in GP
1I need the root answer ...Aragon u r right ..the gap becomes smaller and smaller ...but does it become zero ?? :D ...never..In fact if u look at the progression the gap becomes zero after infinite steps..if u want i will do the math for u :D.
what's wrong then? There is a classic mistake in this approach.
1I need the root answer ...Aragon u r right ..the gap becomes smaller and smaller ...but does it become zero ?? :D ...never..In fact if u look at the progression the gap becomes zero after infinite steps..if u want i will do the math for u :D.
what's wrong then? There is a classic mistake in this approach.
33Well A can reach B's position at just the time B start moving away...
Just a guess knowing that A will catch up :P