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Let a body start moving with a given velocity. $ $ After 1 hour, another body with a greater velocity than the previous one starts moving.

Now, if we think logically, then when the second body nearly reaches the first body, the first body would have moved some more distance during that time interval. So, the body with less velocity will always have first position, but in practical life it doesn't happens.

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closed as off-topic by Lord Shark the Unknown, Henrik, Andrés E. Caicedo, Stella Biderman, user223391 May 1 '17 at 19:43

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – Lord Shark the Unknown, Henrik, Andrés E. Caicedo, Stella Biderman, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ you can't stop time for one body, without stopping it for the other. When the 2nd body reachest the first body, and the first body moves some distance forward (now we stop time to look at it), the second body ALSO had moved some distance forward (in fact it moves more). There is no paradox here $\endgroup$ – frogeyedpeas May 1 '17 at 8:40
  • $\begingroup$ Today, we know it is not necessary to analyze this scenario in terms of every decreasing intervals. The problem in Zeno's day was that the philosophers had no notion of measuring the speed of an object. Today, the solution is a trivial application of the speed-distance-time formula as taught to all school children. $\endgroup$ – Dan Christensen May 22 '17 at 3:13
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Let’s assume two particles ($A$ and $B$) in a straight line motion. Also, $A$ moves faster than $B$.
Then on any given interval of time $t$, $ $ $A$ will move a further distance than $B$.

Let’s take the interval of time to be 1 sec, and say $B$ starts the motion further ahead on the line than $A$. $ $ Then depending on how further ahead $B$ started, and how much faster $A$ is moving, $A$ will not necessarily catch up to $B$ after only one interval of 1 sec.

$A$ will nonetheless move a greater distance than $B$ during this 1 sec. $ $ Let the additional distance moved by $A$ be $x$. $ $ Then after each interval of 1 sec, $A$ will move a distance of $x$ more than $B$ (assuming the speeds involved are all constant).
The additional distances $x$, occurring after consecutive steps of motion (after each 1 sec, in this case) will add up. So after 10 steps (or 10 sec), the additional distance moved by $A$ will be $10x$, after 20 steps (or 20 sec), the additional distance will be $20x$, and after $N$ steps, we’ll have $Nx$.

Now, remember that $B$ started the motion further ahead, and so initially $B$ was some finite distance $y$ away from $A$.
Since $y$ is finite, then given enough steps $N$, $Nx$ will be great enough to equal $y$, and then $Nx>y$, which means $A$ will eventually catch up to $B$, and move past it.
There’s one crucial point to note here. That is, the time between consecutive steps of motion in this case was always 1 sec. The interval of time between steps was thus constant.

However, this is not true in the Zeno’s paradox. In fact, the interval of time between successive steps is continuously decreased in the Zeno’s paradox so that your timeline ends up being bounded. That is, it will never exceed some given number (some given limit).
For instance, in the Zeno’s paradox, the different steps may occur at time $t=1$, then at time $t=1.5$, then at $t=1.75$, etc… which gives the sequence $ $ {1, 1.5, 1.75, 1.875, …},
which is the same as $ $ {1, $ $ (1+0.5), $ $ (1.5+0.25), $ $ (1.75+0.125), …}.

As you can see, each increment of time we add after each step decreases - it is no longer 1 sec – but starts being 0.5 sec, then 0.25 sec, then 0.125, etc…
This means that as the number of steps tends to infinity, the length of the increment of time will tend to $0$. $ $ This in essence means that the above sequence is in fact bounded (both left and right), and so has an upper limit. In this case, this upper limit happens to be 2, which means the time $t$ will never exceed 2 sec!

So, as long as you consider only times less than $t=2$, $ $ $A$ indeed will never catch up to $B$.
But this is absurd, since in reality the time will exceed 2 sec, and in doing so, $A$ will catch up and move past $B$.

Finally, note that the example we chose here, with the above sequence for the time increment, does also make sense in a more general sense. That is, you can choose any sequence you want, and as long as it follows the Zeno’s paradox principle, that is, as long as the time increments tend to 0 as the number of steps tend to infinity, your time $t$ will have a limit.

This indeed shows that there’s no paradox to begin with.

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