1st eq'N of motion :- relation give the velOcitY acquired by a body in time tacceleration= change velocity/time taKen a= v-u/tat= v-uv=u+at2nd eq'N of motion:- uniformly accelerated motion in straight lines= ut+ 1/2 at^2avg velocity= v+u/2s= (u+v/2) × ts= (u+ u + at/2) × t2u+at × t/2s = ut + 1/2 at^23rd eq'N of motion:- velOcitY acquired by a body in covered a distancev= u+ atsquaring botH side we, getv^2 = [u+at^2] or v^2u^2 + 2uat + a^2 t^2v^2 = u^2 + 2a ( ut + 1/2 at^2)v^2 = u^2 + 2aS ProVed cheer guys hurrrrahhhhhhhhhhhhhhhhhhhhhh::::

Equations of Motion:

1. First Equation of Motion:
- The first equation of motion is: v = u + at
- Where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
- For example, if a car starts from rest (u = 0 m/s) and accelerates at 2 m/s^2 for 5 seconds, the final velocity can be calculated using the equation v = 0 + (2 * 5) = 10 m/s.

2. Second Equation of Motion:
- The second equation of motion is: s = ut + (1/2)at^2
- Where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time taken.
- For instance, if a ball is thrown vertically upwards with an initial velocity of 20 m/s and the acceleration due to gravity is 9.8 m/s^2, the maximum height it reaches can be calculated using the equation s = (20 * t) + (0.5 * 9.8 * t^2).

3. Third Equation of Motion:
- The third equation of motion is: v^2 = u^2 + 2as
- Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
- For example, if a stone is dropped from a height of 100 meters (s = -100 m) with an initial velocity of 0 m/s, the final velocity just before hitting the ground can be calculated using the equation v^2 = 0 + 2(-9.8)(-100).