An object moves with speed $v_{1}, v_{2}$ and $v_{3}$ along a line segment $AB$ , $BC$ and $CD$ respectively as shown in figure. Where $AB = BC$ and $AD = 3 AB$ , then average speed of the object will be:

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

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

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

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

answer is B.

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Detailed Solution

$\begin{array}{l} V_{av} = \frac{Total Distance}{Total Time} = \frac{3 s}{\frac{s}{v_{1}} + \frac{s}{v_{2}} + \frac{s}{v_{3}}} \\ = \frac{3 v_{1} v_{2} v_{3}}{v_{1} v_{2} + v_{2} v_{3} + v_{1} v_{3}} \end{array}$

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