Question

Answer the following.

1. Derive the equation

a. v = u+at

b. s= ut+1/2at2

C. V2=u2+2as​

Answer 1

i hope it helps u buddy!!!

i hope it helps u buddy!!!plz mark it as brainliest

Answer 2

${\bold{\huge{\blue{\underline{\red{An}\pink{sw}\green{er}\purple{:-}}}}}}$

Let a body moves in a straight path with initial velocity u.

Let the acceleration of body be a

V be the velocity of the body of the time t.

Now,

$\bf\boxed\star\red{\underline{\underline{{For\:first\:equation\:of\: motion}}}}$

Slopes of velocity time graph gives acceleration.

Therefore,

Slope of AB = tan θ = $\bf\dfrac{BC}{AC}$

$\bf\:a=\dfrac{v-u}{t}$

$\bf\:v=at+u$

$\bf\orange{{\therefore\:v=u+at}}$

$\bf\boxed\star\pink{\underline{\underline{{For\:second\:equation\:of\: motion}}}}$

Displacement= Area of velocity time graph with time axis

Displacement= Area of rectangle OACM + area of ∆ABC

$\bf\:s=OA\times\:OM+\dfrac{1}{2}\times\:AC\times\:BC$

$\bf\:s=u\times\:t+\dfrac{1}{2}\times\:t\times\:(v-u)$

$\bf\:s=ut+\dfrac{1}{2}\times\:t\times\:(u+at-u)$

$\bf\pink{{s=ut+\dfrac{1}{2}a{t}^{2}}}$

$\bf\boxed\star\green{\underline{\underline{{For\:Third\:equation\:of\: motion}}}}$

Displacement= Area of trapezium OABM

$\bf\:s=\dfrac{1}{2}(OA+BM)\times\:OM$

$\bf\:s=\dfrac{1}{2}(u+v)\times\:t$

$\bf\:s=\dfrac{1}{2}\:(u+v)\times\:\dfrac{v-u}{a}$

$\bf\:2as=(v+u)\:(v-u)$

$\bf\:2as={v}^{2}-{u}^{2}$

$\bf\:{u}^{2}+2as={v}^{2}$

$\bf\blue{{\therefore{v}^{2}={u}^{2}+2as}}$

$\bf\:t=\dfrac{v-u}{a}$ comes from here,

↓

$\bf\:v=u+at$

$\bf\:v-u=at$

$\bf\:at=v-u$

$\bf\:t=\dfrac{v-u}{a}$