Question

derive the equation 2as=vt2-vo2​

Answer 1

We are asked to derive,

⋆ The third equation of motion that is 2as = v²-u² Let's derive it by graphical method!

Firstly, before deriving let us know that what are equations of motion.

⋆ There are three equations of motion. The equations of motion are named as

◆ Velocity time relationship

◆ Position time relationship

◆ Position velocity relationship

⋆ Therefore, velocity time relationship, Position time relationship and Position velocity relationship are the three equations of motion respectively.

◆ Velocity time relationship

That's the equation,

• ${\small{\underline{\boxed{\sf{v \: = u \: + at}}}}}$

◆ Position time relationship

That's the equation,

• ${\small{\underline{\boxed{\sf{s \: = ut \: + \dfrac{1}{2} \times at^2}}}}}$

◆ Position velocity relationship

That's the equation,

• ${\small{\underline{\boxed{\sf{2as \: = v^2 - u^2}}}}}$

(Where, v denotes final velocity , u denotes initial velocity , a denotes acceleration , t denotes time , s denotes displacement or distance)

~ Firstly according to the graph,

⇢ AO = DC = u (Initial velocity)

⇢ AD = OC = t (Time)

⇢ EO = BC = v (Final velocity)

~ Now let's derive!

⇢ From the velocity-tme graph shown in the attachment the distance s travelled by the object in time t moving under uniform acceleration a is given by the area enclosed within the trapezium OABC under the graph.

$:\implies \tt Distance \: = OABC \\ \\ :\implies \tt Distance \: = OABC \: trapeizum \: area \\ \\ :\implies \tt s \: = OABC \: trapeizum \: area \\ \\ :\implies \tt Distance \: = \dfrac{1}{2} \times (Sum \: of \: parallel \: sides) \times Height \\ \\ :\implies \tt s \: = \dfrac{1}{2} \times (a+b) \times h \\ \\ :\implies \tt s \: = \dfrac{1}{2} \times (AO+BC) \times OC \\ \\ :\implies \tt s \: = \dfrac{1}{2} \times (u+v) \times t \\ \\ :\implies \tt s \: = \dfrac{1}{2} \times (u+vt) \\ \\ :\implies \tt s \: = \dfrac{u+vt}{2} \quad {\pmb{\sf{Eq_n \: 1^{st}}}} \\ \\ \sf From \: v-t \: relationship \: we \: get \\ \\ :\implies \tt t \: = \dfrac{v-u}{a} \quad {\pmb{\sf{Eq_n \: 2^{nd}}}} \\ \\ \sf From \: 1st \: \& \: 2nd \: equation \\ \\ :\implies \tt s \: = \dfrac{(v+u) \times (v-u)}{2a} \\ \\ :\implies \tt s \: = \dfrac{v^2 - u^2}{2a} \\ \\ :\implies \tt 2as \: = v^2 - u^2 \\ \\ {\pmb{\sf{Henceforth, \: derived!!!}}}$