Question

If a body covers 3 equal distances with speeds v1,v2,v3 then what is the average speed of body.​

Answer 1

Answer:

Car travels 3 equal distances with speeds v1, v2 , v3 . We need to know that average speed is defined as the ratio of total distance to the total time taken.

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

$= > avg. \: v = \dfrac{s}{ \dfrac{ (\frac{s}{3} )}{v1} + \dfrac{(\frac{s}{3} )}{v2} + \dfrac{(\frac{s}{3} )}{v3} }$

Cancelling "s" on numerator and denominator :

$= > avg. \: v = \dfrac{1}{ \dfrac{ (\frac{1}{3} )}{v1} + \dfrac{(\frac{1}{3} )}{v2} + \dfrac{(\frac{1}{3} )}{v3} }$

$= > avg. \: v = \dfrac{1}{ \dfrac{1}{3(v1)} + \dfrac{1}{3(v2)} + \dfrac{1}{3(v3)} }$

$= > avg. \: v = \dfrac{3(v1)(v2)(v3)}{(v1)(v2) + (v2)(v3) + (v3)(v1)}$

So final answer :

$\boxed{ \red{ \sf{ \bold{avg. \: v = \dfrac{3(v1)(v2)(v3)}{(v1)(v2) + (v2)(v3) + (v3)(v1)} }}}}$

Answer 2

$</p><p>\tt{\huge{\pink{Hello!!! }}}$

$\tt{ \purple{ \implies{ \dfrac{v_{1}v_{2}v_{3}}{v_{1}v _{2}+ v_{2}v_{3} + v_{3}v-{1} }}}$