Velocity after applying a force in the vacuum I’m sorry for so simple question, but I just need to be sure.
I understand, that the changing of the speed occurs only when the force is applied, I understand that if one punch a ball in the free space it will infinitely move with a constant velocity
Some point-like body with mass $m$ is situated in vacuum, and has initial velocity $v_1=0 \space m/s$. 
Some force is now acting on a body for a infinitely short period of the time.
The acceleration that gives the application of this force to body equals $a=5 \space m/s^2$.
The velocity after will be $v_2=0+5 =5\space m/s$?

Also, if the force is acting for a non-infinitely short period of time how to calculate then?
I found this from https://physics.stackexchange.com/a/231120/255554
$$x=( x + \frac{|F| }{2m} t^{2} ) $$
Seems it can be applied for both of my cases, but I don’t know why there is 2 times mass

And, can you, please confirm, if 1 Newton is the force that during 1 second changes the 1 kg body velocity on 1 m/s, then 2 Newtons is the force that changes:


*

*if mass is same: during 1 second velocity on 2 m/s

*if mass is 2 kg: during 1 second velocity on 1 m/s


Am I understanding correctly?
 A: In order to determine the velocity as a result of the application of the force, you need to know the duration (time) of the application of the force on $m$, or the displacement $x$ of the mass $m$ during the application of the force, and the force as a function of time if not constant.
Your value of $v_2$ is based on a constant acceleration of 5 m/s$^2$ due to a constant force applied for a duration of 1 second, and comes from the equation:
$$v_{f}=v_{i}+at$$
where $v_{i}$ is the initial velocity and $v_f$ is the final velocity
You equation for displacement $x$ is based on the equation
$$x_{f}=x_{i}+\frac{at^2}{2}$$
Where $m_f$ and $m_i$ are the final and initial displacements
Substituting $$a=\frac{F}{m}$$
Gives
$$x_{f}=x_{i}+\frac{F}{2m}t^2$$
Hope this helps.
A: Firstly, the body will only accelerate while the force is being applied, and it will move at a constant velocity the instant the force stops being applied.
Your final equation is just a variation on
$$x=\frac12at^2$$
Why that factor of ½ arises can be shown using elementary calculus, or by a geometrical argument.

Both statements about a 1 Newton force in your update are correct.
A: You appear to be applying the impulse $5 m Ns$, where $m$ is the mass (in kg), which leads to the applied force $$F(t) = 5 m \delta (t) N,$$
that is, the impulse acts for an infinitesimally small time at the time $t = 0$.
Applying Newton's Law, $F = m a$ with $a = \frac{d v}{d t} = \frac{d^2 x}{d t^2}$ yields $$\frac{ d v}{d t} = 5 \delta (t).$$
This can be easily integrated to give $$ \int_{v = 0}^{v(t)} dv = \int_{t' = - \infty}^{t'= t} 5 \delta (t') dt,$$
since for times less than $t = 0$ the particle is at rest and for $t > 0$ the particle travels with speed $v$. This gives $v (t) = 5 m/s$ for times $t > 0$. So you are correct, the speed after the impulse is a constant $5 m/s$.
Note that the mass $m$ cancels out in the problem as specified.
For the more general case when the force has the form $F = G(t)$ then have the result
$$ m \int_{v = 0}^{v(t)} dv = \int_{t' = - \infty}^{t' = t} G (t') dt,$$
again assuming that the particle is at rest in the far past.
