Question

how to derive 3rd equation of motion v^2-u^2=2as​

Answer 1

$\bold\red{distance \: = average \: v. \: \times time}$

$\bold\green{ s = \frac{v + u}{2} \: \times t}$

$\bold\green{t = \frac{v - u}{a}} \: \\ \\ put \: it \: into \: above \: eq.$

we get,

$\bold\green{ s = \frac{v + u}{2} \times \frac{v - u}{a} }$

$\bold\green{ \: s = \frac{ {v}^{2} - {u}^{2} }{2a} }$

$\bold\green{ 2as \: = {v}^{2} - {u}^{2} }$

$\huge\bold\red{proved}$

$\huge\bold\green{I \: hope \: it \: helps}$

Answer 2

From 1st equation of motion
V= U +AT
Therefore V-U = AT

We know that
S=(U+V/2) ×T
S/T = U +V/2
U+V = 2S/T
Finally we get
(V-U) (V+U) = AT × 2S/T
Therefore
V^2 - U^2 = 2as