Derivation of the kinematic equations
[ I have posted a youtube video on derving the kinematic equations, here is the link: deriving the 1-D kinematic equations. ]We start with the definitions of average acceleration, and average velocity,
$\bar a=\dfrac{\Delta v}{\Delta t}$$\bar v=\dfrac{\Delta x}{\Delta t}$
Kinematic equations are derived with the assumption that acceleration is constant. When the acceleration is constant, average and instantaneous acceleration are the same. So, we can replace $\bar a$ with $a$.
In the notations, we do some changes, first, we take the initial time, $t_1$ as 0, and the final time, $t_2$ as $t$, so we have the time interval, $\Delta t=t_2-t_1=t-0=t$.
And we take the initial velocity, $v_1=v_0$, and the final velocity, $v_2=v$. So $v$ is now the velocity after the time interval of $t$, or the velocity at time $t$. The change in velocity is therefore, $\Delta v=v_2-v_1=v-v_0$.
Putting these in the average acceleration equation:$a=\dfrac{v-v_0}{t}$.
Solving for $v$,
$\boxed{v=v_0+at}$
This is the first kinematic equation.
Next, take the average velocity equation, and substitute $\Delta t =t$, you will get,
$\bar v=\dfrac{\Delta x}{t}$.
From this,
$\Delta x=\bar v t$
We have two velocities, the initial velocity, $v_0$ and final velocity, $v$. If you add them and divide by 2, you get another equation for average velocity:$ \bar {v} = \dfrac{v_0 + v}{2} $
Substituting this into the above equation,
$\boxed{\Delta x=\dfrac{1}{2}(v_0+v)t}$.
This is the second kinematic equation.
Now, substitute, $v$ from the first kinematic equation in the second kinematic equation,
$\Delta x=\dfrac{1}{2}(v_0+v_0+at)t$.
Simplifying,
$\boxed{\Delta x=v_0t+\dfrac{1}{2}at^2}$.
This is the third kinematic equation.
Now, take the first kinematic equation, and solve for $t$, you will get,
$t = \dfrac{v-v_0}{a}$
Substitute this $t$ in the second kinematic equation, you will get
$\Delta x=\dfrac{1}{2}(v_0+v)\, \dfrac{v-v_0}{a}$.
$=\dfrac{1}{2a}(v^2-v_0^2)$.
[Here I have used the identity, $(A+B)(A-B)=A^2-B^2]$
Solving for $v^2$,$\boxed{v^2 = v_0^2 + 2a \Delta x}$
This is the fourth kinematic equation.