Ok so your understanding is absolutely correct... while people have given quite a few good answers... I would like to present you my point of view...

$ acceleration = \frac{dv}{dt} $  
now $\frac{dv}{dt} $ can be represented as $ \frac{dv}{dx} \frac{dx}{dt} $                
$\frac{dx}{dt} = V $                       
Thereforce $a = v \frac{dv}{dx}$               
$ a dx = v dv $  
of intergarting with resepect to displacement we get           
$v^{2} - u^{2} = 2 a s $  where $s$ is the displacement
Multiply both sides by mass $m$  
$mv^{2} - mu^{2} = 2 m a s$  
$\frac{1}{2} mv^{2} - \frac{1}{2} mu^{2} = m a s$        
$m a s$ is now defined as work $W$
and $\frac{1}{2} mv^{2} - \frac{1}{2} mu^{2}$ is defined as change in kinetic energy KE. This is the derivation of the famous work energy theorem.....I hope this helps you