Question

A moving body travels distance s with constant speed v₁ and travels distance s with uniform speed v₂. prove that his average speed is 2v₁v₂/v₁+v₂.

Answer 1

Answer:

hope my answer helps you.........

Answer 2

Answer:

$\textrm{average\ speed}\ =\ \dfrac{2.v1.v2}{v1+v2}$

Explanation:

Given,

distance moved by body in first interval= s

speed of body in first interval = v1

Since,

$\textrm{time\ taken,t1}\ =\ \dfrac{distance}{speed}\\\\\ =\ \dfrac{s}{v1}$

distance moved by body in second interval = s

speed of body in second interval = v2

Since,

$\textrm{time\ taken,t2}\ =\ \dfrac{distance}{speed}\\\\\ =\ \dfrac{s}{v2}$

Now,

$\textrm{Average\ speed}\ =\ \dfrac{total\ distance\ covered}{total\ time\ taken}\\\\\ =\ \dfrac{s\ +\ s}{\dfrac{s}{v1}+\dfrac{s}{v2}}\ =\ \dfrac{2.v1.v2}{v1+v2}$

So, the average speed of the body is $\dfrac{2.v1.v2}{v1+v2}$