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Assuming the spacetime principle, if the space is modified the time does too. So if the velocity in the space is increase, does the time slow down? What happens if the speed is the speed of light, does the time stops?

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No, for several reasons.

First, the idea of time "slowing down" is a little bit of a misnomer. If you were traveling at relativistic speeds, you would not perceive the passage of time any differently than you do right now. It's only when you compare your clocks to an observer in another reference frame (let's me, sitting in my living room, at rest with respect to the ground) that you would notice that your clock shows a smaller passage of time than mine does. So within your own frame of reference, it makes no sense to talk about time ticking slow or fast, it's only when you compare your clock to a clock in another reference frame that you can compare passages in time.

Also, and this is probably the larger hang-up to your question, is that a particle with mass can never reach the speed of light. With truly astounding amounts of energy, you could get arbitrarily close to the speed of light, but never reach it. The reason is because at relativistic speeds, the equation for kinetic energy is different than the classical equation for kinetic energy that you're used to.

Classically, the kinetic energy of a moving particle is given by $$KE=\frac{1}{2}mv^2$$ where we can reach any finite velocity that we want, so long as we do enough work on the object to increase its kinetic energy to the proper amount. However, it turns out that this is equation is only an approximation for small velocities. Relativistically, the equation for kinetic energy is $$KE=\frac{mc^2}{\sqrt{1-\frac{v}{c}^2}}-mc^2$$. What this equation shows you is that for any finite amount of kinetic energy on an object $v<c$. If you have trouble seeing this, attack the equation from a different angle. Let's say I'm crazy, and it is possible to travel at the speed of light. So let's take $v=c$ and substitute that into our relativistic kinetic energy equation. Oh no! You'll see that we get a divide by zero error, because our entire denominator goes to zero. This means we would need an infinite amount of kinetic energy for our object and that's just not possible.

So I hate to burst your bubble, but no light-speed travel for you anytime soon. (Or ever).

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Its not possible to stop time but using relativity it can be thought of to be slowed down .
Nothing can be faster than the speed of light so its not possible .
Even when we near it , energy tends to become infinity .

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Things will be bigger/smaller/slower with speed as $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$.

Thus, as $v \rightarrow c$,

  1. Mass (or energy) goes to infinite,
  2. Time goes to zero.

But it is only a limit, which is unreachable (at least in the special relativity), because of (1).

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    $\begingroup$ This is reachable for photons however, since those are already massless. $\endgroup$
    – DK2AX
    Commented Mar 5, 2015 at 11:03
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    $\begingroup$ This is a very incomplete answer. For example: What is $1/\sqrt{1-v^2/c^2}$? Where did it come from? What "things" do you mean? How does mass/energy go to $\infty$? How does time go to 0? $\endgroup$
    – Kyle Kanos
    Commented Mar 5, 2015 at 15:21
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    $\begingroup$ @KyleKanos $v$, $c$, $\infty$ and such have their common meanings. "Things" means here mass points moving slower than light. Thank you. $\endgroup$
    – peterh
    Commented Mar 5, 2015 at 23:46
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    $\begingroup$ @KyleKanos It was a simple question, which deserved a simple answer. I couldn't include a 200 page long book with the axiomatical description of the special relativity. I don't really understand, what is not clear to you. You know also SR, probably better as me. But this was exactly (the Lorentz-transformation of the spacetime coordinates) which couldn't been written there, because the question required clearly the possible simplest answer. $\endgroup$
    – peterh
    Commented Mar 6, 2015 at 4:04
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    $\begingroup$ I'm also not suggesting that I don't understand your post. I know what you mean to say, but you are doing a terrible job at actually saying it. What I am saying is that John Q Public who doesn't have any physics knowledge would look at your post would have no clue what you were talking about. $\endgroup$
    – Kyle Kanos
    Commented Mar 6, 2015 at 4:12

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