Question

Derive V2 = Vo

2 + 2ax using v-t graph​

Answer 1

Explanation:

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Answer 2

Derivation of v² = u² + 2ax :

First of all, see the graph carefully :

• The initial velocity is u , final velocity is v. The acceleration is a and displacement is x.

Now, displacement is given by area under v-t graph :

$x = area \: of \: trapezium$

$\implies x = \dfrac{1}{2} \times (u + v) \times t$

• Now, 't' can be written as [(v-u)/a].

$\implies x = \dfrac{1}{2} \times (u + v) \times \dfrac{v - u}{a}$

$\implies x = \dfrac{1}{2} \times (v + u) \times \dfrac{v - u}{a}$

$\implies x = \dfrac{1}{2} \times \dfrac{ {v}^{2} - {u}^{2} }{a}$

$\implies x = \dfrac{ {v}^{2} - {u}^{2} }{2a}$

$\boxed{ \implies {v}^{2} = {u}^{2} + 2ax}$

[Hence Proved].