Question

Prove it :-
V^2= U^2 + 2as



U=Initial velocity
a=aceelaration
T=Time
V=Final velocity
S=Dispalcement​

Answer 1

By definition,

$\rm Acceleration = \dfrac{Change \ in \ velocity}{Time \ Taken}$

$\rm \dashrightarrow a= \dfrac{Final \ velocity-Initial \ velocity}{Time \ Taken}$

$\rm \dashrightarrow a = \dfrac{v-u}{t}$

$\rm \dashrightarrow at = v - u$

$\bf \dashrightarrow v = u+at \qquad\qquad ..[1]$

Also,

Displacement = Average velocity × time

or

$\rm S = \dfrac{v+u}{2} \times t$

From [1] , $\rm v = u + at$

or

$\rm t = \dfrac{v-u}{a}$

$\therefore \qquad \rm S = \dfrac{v+u}{2} \times \dfrac{v-u}{a}$

$\rm\dashrightarrow S = \dfrac{(v+u)(v-u)}{2a}$

We know that,

(a+b)(a-b) = a² - b²

Hence,

$\rm\dashrightarrow S = \dfrac{v^2-u^2}{2a}$

$\bf\dashrightarrow v^2-u^2=2as$

Hence, proved.

Answer 2

Answer:

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