Basic question about acceleration [duplicate]

This question already has an answer here:

Very basic question.

Please show where I'm wrong in the following reasoning.

The movement of an object in function of time could be described as $$x(t) = v t + x_{i}$$ if velocity is constant.

If velocity is not constant then $$x(t) = v(t)\cdot t + x_{i}$$ where $$v(t) = a t + v_{i}$$ with a being constant.

Now if I substitute $v(t)$ in $x(t)$ it results $$x(t) = at^2 + v_it + x_i$$

But the general equation for an accelerated object is $$x(t) = \frac{1}{2}at^2 + v_it + x_i$$

Where does the $\frac{1}{2}$ come from?

marked as duplicate by ACuriousMind♦, yuggib, Mike, Kyle Kanos, Qmechanic♦May 30 '15 at 19:24

When the velocity is not constant you have: $$x(t)=\overline{v(t)} t + x_i$$ where $\overline{v(t)}$ is the average velocity from $0$ to $t$. When you have constant acceleration the average velocity is $$\overline{v(t)}=\frac{v(0)+v(t)}{2}=\frac{at}{2} + v_i$$ which will give the correct result.
$$v(t)=\int_{\tau=0}^t a(\tau)\;d\tau+v_i\\ x(t)=\int_{\tau=0}^t v(\tau)\;d\tau+x_i$$
That is the average velocity is: $$\overline{v(t)}=\frac{\int_{\tau=0}^t v(\tau)\;d\tau}{t}$$
In order to express the position as a function of the velocity you have to integrate with respect to time. When the velocity is constant this integral is simple, namely $vt+C$. However once the velocity becomes a function of time this integral will change and will in general not be equal to $v(t)t+C$. You actually have to integrate $v(t)$ with respect to $t$ in order to find the position as a function of time.
The place you are wrong is in supposing $x(t) = v(t).t + x_i$ when acceleration is constant. You will have to derive the relation using integration. The above relation only holds when velocity is constant, but when velocity varies(i.e acceleration is non zero), this relation is no longer true.