Why is it the case that KE = 0.5mv^2
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,
$$\begin{align}
E_\text{k}&amp;=W\\
&amp;=Fs\\
&amp;=mas\\
&amp;=ma\cdot\frac{1}{2}vt=ma\cdot\frac{1}{2}at^2=\frac{1}{2}m(at)^2\\
&amp;=\frac{1}{2}mv^2
\end{align}$$