# Derivation of three Equations of Motion (Mathematical Method)

Let an object move with initial velocity (u) and with (t) time the velocity became (v) than:-

**Acceleration:- **The rate of change in velocity known as acceleration.

a = v – u/t

at = v – u

v = u + at. Equation 1

**Displacement:- ** Average velocity x time

S = v + u/2 x t

u + at + u/2 x t

S = 2ut + at^2/2

S = ut + 1/2at^2. Equation 2

**Squaring equation 1**

V^2 = ( u + at)^2

V^2 = u^2 + at^2 + 2at

v^2 = u^2 + 2a (ut + 1/2at^2)

V^2 = u^2 + 2as Third equation

**Quation related to the equations **

A car is moving in 20/sec velocity and suddenly it applies the break and stop the car after 20m distance than find the time taken by acr to stop it?

From equation third

V^2 = u^2 + 2as

(0)^2 = 20^2 + 2a.20

0 = 400 + 40a

a = -400/40

a = -10

V = u + at

0 = 20 + (-10)t

10t = 20

t = 20/10

t = 2sec

A stone through upward with the velocity of 10m/sec and after 6 second it reaches in hands of man. Find the height covered by the stone?

v = ut + 1/2 at^2

10×3 + 1/2 (-10)x3^2

30x-45

-15 ( height cannot be negative )

15 answer