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Derive the equations of uniformly accelerated motion by graphical method.
Solution
Now at $t=0$ the velocity of the particle be $v_{0}$ and at $t=t$ the velocity of the particle be $v$.
$\therefore$ Acceleration $a=$ slope of line $\mathrm{AB}$
$=\frac{v-v_{0}}{t-0}$
$\therefore a =\frac{v-v_{0}}{t}$
$\therefore v =v_{0}+a t$
Now displacement of a particle during time $t$ is equal to the area of region OABCD in $v \rightarrow t$ graph.
$\therefore x =\text { Area of rectangle OACD }+\text { Area of } \Delta \mathrm{ACB}$
$=v_{0} t+\frac{1}{2}\left(v-v_{0}\right) t$
$\therefore x=v_{0} t+\frac{1}{2} a t^{2}$
but average velocity,
$x=\left(\frac{v+v_{0}}{2}\right) t$
For, equation $( 3 ) $ and equation $(1)$,
$t=\frac{v-v_{0}}{a}$
$\therefore x=\left(\frac{v+v_{0}}{z}\right)\left(\frac{v-v_{0}}{a}\right)$
$\therefore v^{2}-v_{0}^{2}=2 a x$
Now at $t=0$ the velocity of the particle be $v_{0}$ and at $t=t$ the velocity of the particle be $v$.
