Ok so your understanding is absolutely correct...while while people have given quite a few good answers...i I would like to present you my point of view...
acceleration=dv/dt$ acceleration = \frac{dv}{dt} $
now dv/dt$\frac{dv}{dt} $ can be represented as (dv/dx)*(dx/dt)$ \frac{dv}{dx} \frac{dx}{dt} $
dx/dt=V$\frac{dx}{dt} = V $
Thereforce
a=vdv/dx $a = v \frac{dv}{dx}$
adx=vdv$ a dx = v dv $
of intergarting with resepect to displacement we get
v²-u²=2as$v^{2} - u^{2} = 2 a s $ where s$s$ is the displacement
Multiply both sides by mass 'm'$m$
mv²-mu²=2mas$mv^{2} - mu^{2} = 2 m a s$
1/2mv²-1/2mu²=mas$\frac{1}{2} mv^{2} - \frac{1}{2} mu^{2} = m a s$
mas$m a s$ is now defined as work 'W'$W$
and 1/2mv²-1/2mu²$\frac{1}{2} mv^{2} - \frac{1}{2} mu^{2}$ is defined as change in kinetic energy KE
This. This is the derviationderivation of the famous work energy theorem.....I hope this helps you
