Question

Prove 2 a s =v×v -u×u
Please don't copy from Google..
I will mark brainalist....

Answer 1

We know acc=vdv/dx
Doing integration
Adx =vdv
Limit 0 to s in Lhs and in rhs u to V
U= initial and v= final velocity
As=[v^2 /2] limit from u to v
As=( v^2 - u^2) /2
2as= v^2 - u^2

Answer 2

From the velocity time graph the distance s travelled by the object in time t moving under uniform acceleration a is given by te area enclosed within the trapezium OABC under the graph that is

S= area of the trapezium

= 1/2 × h × (sum of || sides)

= OA+BC×OC/2

substituting

OA=u, BC=v , OC=t

We get

S = u+v×t/2

From 1st eq. Of motion

We get

V=u+at

t=v-u/a

Using both equations we have

S=v+u v-u/2a

2as=v square- u square

Hope it helps