Question

if a body moved s1 with speed v1 and distance s2 with speed v2. show that its average speed is:Vav=v1v2(s1+s2/s1v2+s2v1)​

Answer 1

Given:

A body moved s1 with speed v1 and distance s2 with speed v2.

To find:

Average speed is Vav=v1v2(s1+s2/s1v2+s2v1)

Calculation:

Let average velocity be v_(av)

$\rm\therefore \: v_{av} = \dfrac{total \: displacement}{total \: time}$

$\rm\implies\: v_{av} = \dfrac{s_{1} + s_{2}}{ t_{1} + t_{2}}$

$\rm\implies\: v_{av} = \dfrac{s_{1} + s_{2}}{ ( \frac{s_{1}}{v_{1}}) + (\frac{s_{2}}{v_{2}}) }$

$\rm\implies\: v_{av} = \dfrac{s_{1} + s_{2}}{ ( \frac{s_{1}v_{2} +s_{2}v_{1} }{v_{1}v_{2}}) }$

$\rm\implies\: v_{av} = \dfrac{(s_{1} + s_{2})(v_{1}v_{2})}{ ( s_{1}v_{2} +s_{2}v_{1}) }$

$\rm\implies\: v_{av} = ( v_{1}v_{2}) \times \bigg(\dfrac{s_{1} + s_{2}}{ s_{1}v_{2} +s_{2}v_{1} } \bigg)$

So, final answer is :

$\boxed{ \bf\: v_{av} = ( v_{1}v_{2}) \times \bigg(\dfrac{s_{1} + s_{2}}{ s_{1}v_{2} +s_{2}v_{1} } \bigg)}$