Derive three equation of motion by …

We have the third equation of motion as,
v² = U² + 2aS
Where,
v is the final velocity,
S is said to be as displacement of the particle,
u is the initial velocity,
a = acceleration (constant),
At t = 0 sec, u is the initial velocity of the particle.
At t = t sec, v is the final velocity of the particle.
Now, if we draw the graph for the above set of statements then, from the graph we have,
(Displacement)S = Area of Rectangle + Area of Triangle
{tex}S=\frac{u t+1}{2(v-u)(t)}{/tex} .........eqn1
From the first equation of motion we have {tex}t=\frac {(v-u)}{a}{/tex}
Now let us substitute the value of t in eqn 1
{tex}S=u\left(\frac{v-u}{a}\right)+\frac{\frac{1}{2}((v-u)(v-u))}{a}{/tex}
By simplifying and rearranging the terms,
We get the expression as
v² = u² + 2aS = third equation of motion.
Hence the third equation of motion is proved with the help of graphical method.