Question

derive the third equation of motion ​

Answer 1

Third equation of motion -:

=> Distance travelled = Average Velocity × Time

=> Distance travelled =InitialVelocity+FinalVelocity2×Time

=> s= v+u2×t

From equation (i) => v=u+at

or t=v−ua

∴ s= u+v2×v−ua = v²−u²2a

or v²-u² = 2as

or v²= u²+2as ——————(iii)

Where ,

u =initial velocity

v=final velocity

t= time

s= distance

Answer 2

Third equation of motion

From equation (1) we have-

$v=u+at\\\\t=\bigg(\dfrac{v-u}{a}\bigg)$

From equation (2) we have -

$S=ut+\dfrac{1}{2}\:at^2$

Put the value of 't' in equation (2)

We get,

$S=u\bigg(\dfrac{v-u}{a}\bigg)+\dfrac{1}{2}\:a\bigg(\dfrac{v-u}{a}\bigg)^2\\\\\\S=\dfrac{uv-u^2}{a}+\dfrac{a(v^2+u^2-2u\:v)}{2\:a^2}\\\\\\S=\dfrac{uv-u^2}{a}+\dfrac{(v^2+u^2-2u\:v)}{2a}\\\\\\S=\dfrac{2u\:v-2u^2+v^2+u^2-2u\:v}{2a}\\\\\\2as=v^2-u^2\\\\\\v^2=u^2+2as.............(3)\:eq.\:of\:motion.$