Equation of distance and time 
How is this equation derived?
  $$r = r_0 + ut + at²/2$$
  where $r_0$ is the initial position of particle and $r$ is the position of the particle after all the motion it has undergone, $a$ and $t$ have their usual meanings of acceleration and time.

What I think of it is... that this should go on indefinitely like $r = r_0 + ut + at²/2+ jt³/3 +.....$ where j is 4th derivative of displacement. And $jt³/*3*$ because, following the pattern, the denominator is same as the power of time. I'm not sure about. Can you tell me something about it?
 A: It is derived by solving the differential equation
$$\frac{d^2r}{dt^2}=a$$
for constant acceleration $a$; $r_0$ and $u$ are constants of integration.
If the acceleration is not constant, then you can get higher-order terms in $r$, such as one proportional to $t^3$.
Conversely, suppose you had a higher-order term like $t^3$. Then, when you computed the acceleration by differentiating twice, you wouldn’t get a constant.
A: The formula that you stated comes under the assumption a constant force.  Then you have equations of motion
$$ \array{ \dot{r}(t) & = & u(t) \\ \dot{u}(t) & = & a } $$
where $a = F/m$ is constant, $F$ is the force, and $m$ is the mass of your object.  Integrating those equations leads to the polynomial solution that you gave in the question.
If the force were not constant, then you would effectively get higher order terms, as you seemed have thought, although, depending on the situation, a Taylor series like you wrote might not be the most natural way to express the solution, e.g. if it admitted a solution in sine and cosine as for harmonic motion.  Unlike what you wrote, however, the coefficients in any Taylor series that you do write won't follow such a simple, problem-independent pattern, including that the coefficients on the $t$ and $t^2$ terms may differ.
A: If an object is moving with constant acceleration $a(t)=a$ then, it´s velocity is given by the integral of such acceleration:
$v(t)=\int a(t') dt'=\int a dt'$
$v(t)=a t + v_0$
That is, its instantaneous velocity is its acceleration multiplied by the time elapsed plus an inicial velocity.
The displacement of the object will be the integral of its velocity:
$x(t)=\int v(t') dt'=\int (a t' + v_0) dt'$
$x(t)= \frac{at^2}{2}+v_0t+x_0$
Which is the formula you've given. 
A: Start from the definition of acceleration:
$$\tag{1} \mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t) \label{1}$$
In order to get the velocity, we need to integrate the expression from eq. \ref{1}:
$$\tag{2} \mathbf{v}(t)=\int\mathbf{a}(t)dt=\mathbf{a}\cdot t+\mathbf{v}_{0} \label{2}$$
But the velocity is just the derivative wrt to time of the position vector:
$$\tag{3} \mathbf{v}(t)=\frac{d}{dt}\mathbf{r}(t) \label{3}$$
The position vector can then be obtained by integrating the velocity from eq. \ref{2}
$$\tag{4} \mathbf{r}(t)=\int\mathbf{v}(t)dt=\int\left(\mathbf{a}\cdot t+\mathbf{v}_{0}\right)dt=\frac{\mathbf{a}\cdot t^2}{2}+\mathbf{v}_{0}\cdot t+\mathbf{r}_{0} \label{4}$$
A: 

Elementary before learning Calculus.
