Question

How to derive the kinematic equations?

Answer 1

First equation of motion relates velocity, time and acceleration. Now in $\Delta uxy$,
$tan\theta =\frac{xy}{uy}tan\theta =\frac{v-u}{t}$
We also know that tanθ is nothing but the slope and slope of v – t graph represents acceleration.
$\Rightarrow v=u+at$ -------- (1)
This is the first equation of motion where,
v = final velocity
u = initial velocity
a = acceleration
t = time taken
Now coming to the second equation of motion, it relates displacement, velocity, acceleration and time. Area under v – t graph represents the displacement of the body.
In this case,
Displacement = Area of trapezium (ouxt)
$S=\frac{1}{2}\times sumofparallelsides\times height$
$S=\frac{1}{2}\times x(v+u)\times t$ -------- (2)
We can substitute v in terms of others and get final equation as:
$S=ut+\frac{1}{2}a{t}^2$
Where symbols have their usual meaning.
The third equation of motion relates velocity, displacement and acceleration. Using same equation (2),
$S=\frac{1}{2}\times (v+u)\times t$
Using equation (1) if we replace t we get,
$S=\frac{1}{2}\times (v+u)\times \frac{(v-u)}{a}S=\frac{(v^2-u^2)}{2a}{v}^2=u^2+2as$
The above equation represents third equation of motion.