Question

derive the second and third equation of motion graphically​

Answer 1

$\bold{Second\:equation\:of\:motion:-}$

$s = ut + \frac{1}{2} {at}^{2}$

=Distance travelled by object= area of trapezium (OABC)

=Area of OADC ( rectangle)+ area of ABD (triangle)

$= >oa \times ad + \frac{1}{2} \times ab \times bd \\ \\ = > u \times t + \frac{1}{2} \times t \times (v - u) \\ \\ = > ut + \frac{1}{2} \times t \times at \\ \\ s = ut + \frac{1}{2} {at}^{2}$

$\bold{Third\:equation\:of\:motion}$

${v}^{2} = {u}^{2} + 2as$
S= Area of OABC (trapezium)

$s = \frac{(oa + bc) \times oc}{2} \\ \\ s = \frac{(u + v) \times t}{2 } \\ \\ s = \frac{(u + v)}{2} \times \frac{( u+ v)}{a} \\ \\ s = \frac{ {v}^{2} - {u}^{2} }{2a} \\ \\ {v}^{2} = {u}^{2} + 2as$