Question

Derive third equation of motion with help v-t graph​

Answer 1

Suppose a body had a velocity u initially at time 0 seconds and attains a velocity v after t seconds due to a uniform acceleration a.

So its velocity - time graph will be a straight line as follows.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\vector(1,0){50}}\put(0,0){\vector(0,1){50}}\put(0,-5){\sf{0}}\put(-4,10){\sf{u}}\put(-4,35){\sf{v}}\put(25,-5){\sf{t}}\put(0,10){\line(1,1){25}}\multiput(0,35)(2,0){13}{\line(1,0){1}}\multiput(25,0)(0,2){18}{\line(0,1){1}}\end{picture}$

The acceleration is given by the slope of the graph,

$\sf{\longrightarrow a=\dfrac{v-u}{t-0}}$

$\sf{\longrightarrow t=\dfrac{v-u}{a}\quad\quad\dots(1)}$

And the displacement is given by the area under the graph.

Here the region under the graph is a trapezium. So,

$\sf{\longrightarrow s=\dfrac{1}{2}\,t(v+u)}$

From (1),

$\sf{\longrightarrow s=\dfrac{(v+u)(v-u)}{2a}}$

$\sf{\longrightarrow s=\dfrac{v^2-u^2}{2a}}$

$\sf{\longrightarrow\underline{\underline{v^2=u^2+2as}}}$

This is the third equation of motion. Hence derived!