Question

The formula of average velocity is initial velocity + final velocity divided by 2 then what is it's mathematical formulation ​

Answer 1

Given:

Object moves with constant acceleration. Its initial velocity be u and final Velocity be v

To prove:

Its average Velocity is $\dfrac{u+v}{2}$

Proof :

We know that average velocity is the ratio of total displacement to the total time taken

$v \: avg. = \dfrac{total \: displacement}{total \: time}$

$= > v \: avg. = \dfrac{s}{t}$

$= > v \: avg. = \dfrac{ut + \frac{1}{2} a {t}^{2} }{t}$

$= > v \: avg. = u + \frac{1}{2} at$

$= > v \: avg. = u + \dfrac{1}{2} \times ( \dfrac{v - u}{t}) \times t$

$= > v \: avg. = u + \dfrac{v}{2} - \dfrac{u}{2}$

$= > v \: avg. = \dfrac{u}{2} + \dfrac{v}{2}$

$= > v \: avg. = \dfrac{u + v}{2}$

[ Hence Proved ]

Answer 2

We are told that formula for average velocity is sum of Initial velocity (u) and final velocity (v) all over 2.

What we are suppose to do here is to figure the way this formula is formulated.

Basically average velocity is total displacement (s) of a body over total time period (t).

So, this is our basic set up. So let's begin.

Things you must know regrading the question :

• s = ut + ½ at²
• a = (v-u)/t

=> $\sf{Average\:velocity\:=\dfrac{s}{t}}$

$\sf{=>\:Average\: velocity\:=\:\dfrac{ut\:+\:1\:\div\:2\:at^2}{t}....\big[Using\:the\: second\: kinematics\: equation\big]}$

$\sf{=>\:Average\: velocity\:=\:u\:+\:1\:\div\:2\:at}$

$\sf{=>\: Average\: velocity\:=\:u\:+\:\dfrac{1}{2}\:(\dfrac{v-u}{t})\:\times\:t...\big[\because\:a\:=\dfrac{v-u}{t}\big]}$

$\sf{=>Average\: velocity\:=\:u\:+\:\dfrac{1}{2}\:(v-u)}$

=> $\sf{Average\: velocity\:=\:u\:+\:\dfrac{v-u}{2}}$

=> $\sf{Average\: velocity\:=\:u\:+\:\dfrac{v}{2}\:-\:\dfrac{u}{2}}$

=> $\sf{Average\: velocity\:=\:u\:-\:\dfrac{u}{2}\:+\:\dfrac{v}{2}}$

=> $\sf{Average\: velocity\:=\:\dfrac{2u-u}{2}\:+\dfrac{v}{2}}$

=> $\sf{Average\: velocity\:=\dfrac{u}{2}\:+\:\dfrac{v}{2}}$

=> $\large{\rm{\red{Average\: velocity\:=\:\dfrac{u+v}{2}}}}$