### Theory:

Isaac Newton observed that the motion of an object gives a set of three equations of motion which relates the displacement, velocity, acceleration and time of an object under motion, and these equations are called the equation of motion.

More specifically, the equations of motion describe the behaviour of a physical system as a set of mathematical functions in terms of dynamic variables.

The variables include displacement, velocity, acceleration and time of an object under motion. An object is in motion with initial velocity ($$u$$) attains a final velocity ($$v$$) in time ($$t$$) due to uniform acceleration ($$a$$) while covering certain distance ($$s$$). The three equations of motion can be written as,

1. $v=u+\mathit{at}$

2. $s=\mathit{ut}+\frac{1}{2}a{t}^{2}$

3. ${v}^{2}={u}^{2}+2\mathit{as}$
Graphical method:
Using the graphical method, these three equations of motion can be derived in the following way.

1. Velocity-time relation:

In the given velocity-time graph, the initial velocity ($$u$$) is not equal to zero. Here, the object's initial velocity ($$u$$) at point $$A$$, gradually increases to final velocity ($$v$$) at point $$B$$ in time ($$t$$) with uniform acceleration ($$a$$).

The lines $$BC$$ and $$BE$$ are drawn perpendicular from point $$B$$ to the time and the velocity axes, respectively.

Representation of lines on the graph:
Initial velocity ($$u$$) - $$OA$$
Final velocity ($$v$$) - $$BC$$
Time interval ($$t$$) - $$OC$$

Change in velocity in time interval ($$t$$), $$BD$$ $$=$$ $$BC$$ $$–$$ $$CD$$

$$AD$$ is drawn parallel to $$OC$$. Hence, from the graph, it is observed that

$$BC$$ $$=$$ $$BD$$ $$+$$ $$DC$$ $$=$$ $$BD$$ $$+$$ $$OA$$

Substituting $$BC$$ $$=$$ $$v$$ and $$OA$$ $$=$$ $$u$$,

we get, $$v$$ $$=$$ $$BD$$ $$+$$ $$u$$

$\mathit{BD}=v-u\phantom{\rule{0.147em}{0ex}}\to \left(1\right)$

From the graph, the acceleration ($$a$$) of the object is given by

$=\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\frac{\mathit{BD}}{\mathit{AD}}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{\mathit{BD}}{\mathit{OC}}$

Substituting $$OC$$ $$=$$ $$t$$, we get

$a\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{\mathit{BD}}{t}$

$\mathit{BD}=\mathit{at}\to \left(2\right)$

On equating equations ($$1$$) and ($$2$$) we get,

$v=u+\mathit{at}$