Question

A particle travels half the distance with velocity U. the remaining part of the distance is covered with velocity V for the remaining half time. find the average velocity of the particle for complete motion?​

Answer 1

Answer:

Given:

A particle travels half distance with u and other half with v.

To find:

Average velocity of the particle for the whole motion.

Concept:

Average velocityis always represented as the ratio of total displacement to the total time taken by the particle.

$\: \: \: \: \: \: \: \: \boxed{ \sf{ \red{avg. \: v = \dfrac{total \: distance}{total \: time}}}}$

So we have to represent the numerator and denominator with given data from the question.

Calculation:

Let total distance be s :

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

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

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

So final answer is :

$\: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \large{ \sf{ \blue{ avg. \: v = \dfrac{2uv}{u + v}}}}}$

The most important thing that can be commented from this relationship is that:

Average Velocity in this case is actually the geometric mean (GM) of the Velocity data given in the question.

Answer 2

Answer:

$\large\boxed{\sf{V_{avg.}= \dfrac{2uv}{(u + v)} }}$

Explanation:

Let the total displacement is 'x'

According to Question

Half of the distance is covered with velocity 'u'

.°. time taken in this case = $\dfrac{\frac{x}{2}}{u}$

.°. distance left to be covered = x/2

Again, according to question

Left disance is covered with velocity 'v'

.°. time taken in this case = $\dfrac{\frac{x}{2}}{v}$

To find the average velocity:

• We know that, average velocity is the ratio of total distance covered by total time taken.

$= > V_{avg.} = \dfrac{x}{ \dfrac{ \frac{x}{2} }{u} + \dfrac{ \frac{x}{2} }{v} } \\ \\ = > V_{avg.} = \dfrac{xuv}{ \dfrac{x}{2} (u + v)} \\ \\ = > V_{avg.} = \dfrac{uv}{ \dfrac{1}{2} (u + v)} \\ \\ = >\bold{ V_{avg.} = \dfrac{2uv}{(u + v)}}$