Question

For a uniformaly acceleration body with initial and final velocity as Um/s and Vm/s find the average velocity ​

Answer 1

Answer:

Given:

Body is uniformly accelerated.

Initial velocity = u

Final Velocity = v

To find:

Average Velocity

Concept:

Average Velocity is the ratio of total displacement to the total time.

V avg. = (displacement)/(time)

Calculation:

We will calculate displacement and time separately.

Displacement:

v² = u² + 2as

=> 2as = v² - u²

=> s = (v² - u²)/2as......(1)

Time:

v = u + at

=> at = v - u

=> t = (v - u)/a.........(2)

Hence the average Velocity

V avg. = (1)/(2)

=> V avg. = [(v² - u²)/2a]/[(v - u)/a]

=> V avg. = {(v + u)(v - u)}/{2(v - u)}

=> V avg. =( v + u )/2

So final answer is

$v \: avg. = \frac{(u + v)}{2}$

Answer 2

$\LARGE{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

$\large \tt Given \begin{cases} \sf{Intial \: velocity \: = \: u \: ms^{-1}} \\ \sf{Final \: velocity \: = \: v \: ms^{-1}} \end{cases}$

Use equation,

$\Large \star {\underline{\boxed{\blue{\sf{v^2 \: - \: u^2 \: = \: 2as}}}}}$

We have to find Displacement here,

⇒v² - u² = 2as

⇒s = v² - u²/2a

$\Large \leadsto {\boxed{\sf{s \: = \: \frac{v^2 \: - \: u^2}{2a}}}}$

And we know,

$\Large \star {\underline{\boxed{\blue{\sf{a \: = \: \frac{v \: - \: u}{t}}}}}}$

Here we have to find time

⇒t = v - u/t

$\Large \leadsto {\boxed{\sf{t \: = \: \frac{v \: - \: u}{t}}}}$

Now, we have formula for average velocity,

$\Large \star {\underline{\boxed{\blue{\sf{V_{avg} \: = \: \frac{Displacement}{time}}}}}}$

Put values

$\large \rightarrow \displaystyle\sf{v_{avg} \: = \: \frac{ {v}^{2} \: - \: {u}^{2} }{ \dfrac{2 \cancel{a}}{ \dfrac{v \: - \: u}{ \cancel{a}} } } } \\ \\ \rightarrow \large \sf{ v_{avg} \: = \: \frac{ \cancel{(v \: - \: u)}(v \: + \: u)}{ \dfrac{2}{ \cancel{v \: - \: u } }} } \\ \\ \rightarrow \large \sf{v_{avg} \: = \: \dfrac{v \: + \: u}{2} }$

$\LARGE \implies {\boxed{\boxed{\green{\sf{v_{avg} \: = \: \dfrac{v \: + \: u}{2}}}}}}$