Question

Graphically prove that 2as=vf^2_Vi^2​

Answer 1

GiveN :-

• $\sf Third\ equation\ of\ motion\ which \ is , 2as = v^2- u^2.$

To ProvE :-

• $\sf 2as = v^2 - u^2\:using \:graphical\:method.$

SolutioN :-

For this let us consider a object has an initial Velocity of u , was a acclerating with a , and after time interval t its velocity becomes v .

$\sf ( For \ graph \ refer \ to \ the \ attachment :- )$

$\underline{\red{\sf First \ equation \ of \ motion \ is :- }}$

$\large\qquad\qquad\boxed{\pink{\bf v = u + at }}$

$\tt:\implies v = u + at \\\\\tt:\implies at = v - u \\\\\tt:\green{\implies t = \dfrac{v-u}{a}}$

$\underline{\red{\sf Graph \ of \ the \ motion :-}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(12,12)\put(0,0){\vector(1,0){6}}\put(4,0){\line(0,1){4}}\put(0.001,0){\vector(0,1){6}}\put(0,2){\line(1,0){0.2}}\put(.3,2){\line(1,0){0.2}}\put(.6,2){\line(1,0){0.2}}\put(.9,2){\line(1,0){0.2}}\put(1.2,2){\line(1,0){0.2}}\put(1.5,2){\line(1,0){0.2}}\put(1.8,2){\line(1,0){0.2}}\put(2.1,2){\line(1,0){0.2}}\put(2.4,2){\line(1,0){0.2}}\put(2.7,2){\line(1,0){0.2}}\put(3,2){\line(1,0){0.2}}\put(3.3,2){\line(1,0){0.2}}\put(3.6,2){\line(1,0){0.2}}\put(3.9,2){\line(1,0){0.1}}\put(0,4){\line(1,0){4}}\put(0,2){\line(2,1){4}}\put(0,-0.3){ \tt O }\put(4,-0.3){ \tt A }\put( - 0.3,2){ \tt u }\put( - 0.3,4){ \tt v }\put( 0.2,1.6){ \tt C }\put( 4.2,1.6){ \tt E }\put( 4.2,4){ \tt B }\end{picture}$

$\rule{200}2$

Now , we know that the area of ( v - t ) graph gives Displacement .So , here OCBA is a trapezium .

$\underline{\pink{\sf Area \ of \ trapezium \ OCBA :- }}$

$\tt:\implies ar(trap. \ OCBA) = \dfrac{1}{2}\times ( sum \ of \ parallel \ sides ) \times ( distance \ between \ them ) \\\\\tt:\implies ar( OCBA) = \dfrac{1}{2}\times ( OC + AB) \times CE \\\\\tt:\implies ar( OCBA) = \dfrac{1}{2}\times ( u + v ) \times t \\\\\tt:\implies s = \dfrac{1}{2} \times ( v + u ) \times \bigg\lgroup \dfrac{v-u}{a} \bigg\rgroup \\\\\tt:\implies 2s = \bigg\lgroup \dfrac{(v+u)(v-u)}{a}\bigg\rgroup \\\\\tt:\implies 2as = ( v + u ) ( v - u ) \\\\\orange{\underset{\blue{\bf Hence \:Proved}}{\underbrace{\underline{\boxed{\red{\tt\longmapsto 2as = v^2 - u^2 }}}}}}$