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.
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 sequence of time increments gives rise to a convergent series, 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.