Question

define the expression v square = u square 2as graphically ​

Answer 1

Let , a body has initial velocity 'u' and in time 't' , it reaches a final velocity 'v' .

Here ,

Initial Velocity , u = OA

Final Velocity , v = BD

Time , t = OC


Therefore ,


Distance = ar(AOCB)


$= > s = ar(AOCB)$


$= ar(AOCD) + ar(ADB)$


$= OA \times OC + \frac{1}{2} \times BD \times AD$


$= ut + \frac{1}{2} \times at \times t$


$= ut + \frac{1}{2} a {t}^{2}$



So Distance ,$s = ut + \frac{1}{2}a {t}^{2}$



From first equation of motion we get —


$v = u + at$


$= > t = \frac{ v - u}{a}$



Therefore ,



$s = u( \frac{v - u}{a} ) + \frac{1}{2}a {( \frac{v - u}{a} )}^{2}$


$= > s = \frac{uv - {u}^{2} }{a} + \frac{a}{2} ( \frac{ {v}^{2} - 2uv + {u}^{2} }{ {a}^{2} } )$


$= > s = \frac{uv - {u}^{2} }{a} + \frac{ {v}^{2} - 2uv + {u}^{2} }{2a}$


$= > s = \frac{2uv - 2 {u}^{2} + {v}^{2} - 2uv + {u}^{2} }{2a}$


$= > 2as = {v}^{2} - {u}^{2}$


$= > {v}^{2} = {u}^{2} + 2as$



Note :- Refer to the attachment given above .