Let’s begin from the bases of physics, and with the simple things of it. So, let’s begin with the mechanics topic. We can divide it into 2 parts: Kinematics and Dynamics. The first one is the study of the motion of the particle, the second one is the study of how the particle react against the external forces, and is the study of the interaction about 2 particles. Let’s explain something more:

- Kinematics: When you watch a particle changing position, you can see (imaginary) that it does a line (it can be a straight line or a curvilinear one). That line is called trajectory. With that, someone wasted his time by watching a particle moving, and he saw that there was a link between space and time, and here they did that relation:

**v**= (**x**/t)

where “**x**“ is the space (it’s bold because it’s a vector) and “t” is the time to do that space. Removing the vector thing, we have that:

*v = Δx/Δt -> v = (x(2)-x(1))/(t(2)-t(1))*

This is the average speed. To know the instantaneous speed, you must do the limit of the time that goes to zero:

lim (* x/t) *for

*t*-> 0

But we know that we can also write:

lim (*Δx/Δt*) for *t* ->0

We can see that this is the concept of the derivate of the space divided by the derivate of time, and so we have that result:

*v = (dx/dt)*

This is the instantaneous speed. We can solve this differential equation:

*v*dt = dx -> ∫v dt = ∫dx -> x(2) – x(1) = ∫v dt*

*If the speed is constant, we have the uniform motion*

*x(2) = x(1) + v*(t(2)-t(1))*

And this is the main equation for the uniform rectilinear motion. So, we can see that the speed is the variation of the space over time. From here we can see one thing: if the speed is the variation of the space over time, can we do the variation of the speed over time?

The answer is yes, so we have:

**a** = **v**/t -> a = (v(2)-v(1))/(t(2)-t(1))

that quantity was called acceleration. As before, this is the average acceleration. We can do the same things that we did with the speed, so we have

*a = dv/dt*

Solving that differential equation, you have:

*a dt = dv -> ∫a dt = ∫dv -> v(2) -v(1) = ∫a dt*

*If the acceleration is constant, we have the uniformly accelerated motion:*

*v(2) = v(1) + a*(t(2)-t(1))*

Taking the integral form of the uniform motion, we have

*x(2) – x(1) = ∫(v(1)+a*(t(2)-t(1)))dt*

*this is the motion law of the uniformly accelerated motion*

*x(2) = x(1) + v(t(2)-t(1)) + (a*(t(2)-t(1))^2)/2*

[to be continued...]