Why is it the case that KE = 0.5mv^2

To the best of my knowledge, energy equals work, $\mathrm{E = W}$; work equals force multiplied by distance, $\mathrm{W = Fm}$ ; force equals mass multiplied by acceleration, $\mathrm{F = MA}$; and acceleration equals distance per second squared, $\mathrm{A = m \setminus s^2}$. However, when I substitute those values in for the dimensions of $ \mathrm{W}$, I can't transform the result of those substitutions into $ \mathrm{W= \frac{1}{2}Mv^2}$. Perhaps I've done the algebra incorrectly. 


Does the result of those substitutions transform into $ \mathrm{W= \frac{1}{2}Mv^2}$?


If it doesn't transform into that equation, why doesn't it do so, and why do we use an equation that isn't compatible with our other equations? 




The kinetic energy $E\_\text{k}$ equals the work $W$ done on an object:
$$E\_\text{k}=W$$
The work $W$ is the result of a force $F$ on a distance $s$. If $F$ is constant, the work $W$ is equal to the force $F$ multiplied by the distance $s$:
$$W=Fs$$
The force $F$ on an object is equal to the mass $m$ of that object multiplied by the acceleration $a$:
$$F=ma$$
For uniform acceleration $a$, the final velocity $v$ is
$$v=at$$
and the distance $s$ is
$$s=\frac{1}{2}vt=\frac{1}{2}at^2$$


Therefore,
$$\begin{align}
E\_\text{k}&=W\\
&=Fs\\
&=mas\\
&=ma\cdot\frac{1}{2}vt=ma\cdot\frac{1}{2}at^2=\frac{1}{2}m(at)^2\\
&=\frac{1}{2}mv^2
\end{align}$$


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@Loong's answer is perfect but I just want to show another method which can be used even if the acceleration is **not** uniform.


$$a=\frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt}$$
Hence $$a=v\frac{dv}{dx}$$


Force is mass times acceleration.


Hence $$F=ma=mv \frac{dv}{dx}$$


For a small distance $dx$ moved, the work done by the force is $F \times dx$.


By the Work-Energy Theorem, this is also the gain in kinetic energy.


Hence $$\Delta KE = \int\_{x\_1}^{x\_2} Fdx = \int\_{x\_1}^{x\_2} (mv \frac{dv}{dx}) dx = \int\_{v\_1}^{v\_2}mvdv$$


Hence $$\Delta KE = \frac{1}{2}m{v\_2}^2 - \frac{1}{2}m{v\_1}^2$$


If the body starts at rest $v\_1=0$ and we get the kinetic energy as $\frac{1}{2}mv^2$


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