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