Suppose a body is moving with initial velocity v_i and after a certain time t its velocity becomes v_f then the total distance S covered in time t is given by

$S = v_{av} \times t$

\begin{equation} S = \dfrac{vf + Vi}{2} \times t \end{equation}

From the first equation of motion.                 V_f = V_i + at   

Putting the value of V_f in equation (1).

$\dfrac{vf + at vi}{2} \times t$

$\dfrac{2vi+ at}{2} \times t$

$\dfrac{2vit+ at2}{2} \times t$

$\dfrac{2vit}{2}$ + $\dfrac{at2}{2}$

S = V.t + ½ at²

Second Method (Graphical method):

The second equation of motion:

In the speed-time graph shown in the figure, the total distance S travelled by the body is equal to the total area OABD under the graph. That is

Total distance S = area of (rectangle OACD + triangle ABC)

Area of rectangle OACD = OA x OD

                                           = V_i x t

Area of triangle ABC       = ½ (AC x BC)

                                           = ½ t x at

Since Total area OABD = area of rectangle OACD + area of triangle ABC

Putting values in the above equation, we get

                          S = V_it + ½ t x at

                          S = V_it + ½ at²