If a car covers $2 / 5^{t h}$ of the total distance with $v_{1}$ speed and $3 / 5^{t h}$ distance with $v_{2}$ then average speed is

$\frac{1}{2} \sqrt{v_{1} v_{2}}$

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$\frac{v_{1} + v_{2}}{2}$

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$\frac{2 v_{1} v_{2}}{v_{1} + v_{2}}$

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$\frac{v_{1} v_{2}}{3 v_{1} + 2 v_{2}}$

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Solution

The correct option is D
$\frac{v_{1} v_{2}}{3 v_{1} + 2 v_{2}}$

$A v e r a g e s p e e d = \frac{T o t a l d i s t a n c e t r a v e l l e d}{T o t a l t i m e t a k e n}$
= $\frac{x}{\frac{2 x / 5}{v_{1}} + \frac{3 x / 5}{v_{2}}} = \frac{5 v_{1} v_{2}}{3 v_{1} + 2 v_{2}}$

Suggest Corrections

Similar questions

\frac{2}{5^{t h}}

v_{1}

\frac{3}{5^{t h}}

v_{2}

v_{1}

v_{2}

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