Question

derive the eqn v^2-u^2=2as​

Answer 1

SEE THE ATTACHED FILE.....

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Answer 2

Answer:

BY ALGEBRAIC METHOD..........

from 2nd equation of motion : s = ut + 1/2at^2

and from 1st equation of motion : v = u + at

t = v - u / t

put the value of t in equation 2nd , we get

s = u ( v - u / a ) + 1/2 a ( v - u / a )^2

s = u ( v - u / a ) + 1 /2 a ( v^2 + u^2 - 2.v.u / a^2 )

s = uv - u^2 /a + v^2 + u^2 - 2uv / 2a

[ take LCM ]

s = 2uv - 2u^2 + v^2 + u^2 - 2uv / 2a

s = -u^2 + v^2 / 2a

2as = v^2 - u^2

BY GRAPHICAL METHOD......

s = area of trapezium OABC

s = ( OA + CB ) OC / 2

s = ( u + v ) t / 2

[ put the value of t as v - u / a ]

s = u + v / 2 * ( v - u / a )

s = v^2 - u^2 / 2a

2as = v^2 - u^2

hope it helps................