Question

how we derive vsquare=usquare +2as

Answer 1

With the help of graph attached above...

s=area of trapezium OABC
s=1/2(sum of parallel sides) height
s=1/2(OA+BC)*OC
s=½(u+v)t ......(1)

From first equation of motion:v=u+at
v-u=at
v-u/a=t .......(2)

Substituting (2) in (1)
s=½(v+u)*(v-u/a)
2as=(v+u)(v-u)
2as=v²-u²