Question

If a car covers 2/5th of the total distance with v1 speed and 3/5th distance with v2,then the average speed

Answer 1

average speed is total distance covered/total time taken

Answer 2

Answer:

The average speed is $v_{avg}=\dfrac{5v_{1}v_{2}}{2v_{2}+3v_{3}}$.

Explanation:

Given that,

Speed = $v_{1}$

Speed = $v_{2}$

Let the total distance is D.

Distance $d_{1}=\dfrac{2}{5}\ d$

Distance $d_{2}=\dfrac{3}{5}\ d$

The time will be

$t_{1}=\dfrac{d_{1}}{v_{1}}$

$t_{1}=\dfrac{\dfrac{2}{5}d}{v_{1}}$

$t_{2}=\dfrac{\dfrac{3}{5}d}{v_{2}}$

The total time is

$T = t_{1}+t_{2}$

$T = \dfrac{\dfrac{2}{5}s}{v_{1}}+\dfrac{\dfrac{3}{5}s}{v_{2}}$

The average speed is

$v_{avg}=\dfrac{D}{T}$

$v_{avg}=\dfrac{\dfrac{2}{5}\ d+\dfrac{3}{5}\ d}{\dfrac{\dfrac{2}{5}s}{v_{1}}+\dfrac{\dfrac{3}{5}s}{v_{2}}}$

$v_{avg}=\dfrac{5v_{1}v_{2}}{2v_{2}+3v_{3}}$

Hence, The average speed is $v_{avg}=\dfrac{5v_{1}v_{2}}{2v_{2}+3v_{3}}$.