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(C) Average speed is defined as the ratio of total distance traveled to the total time taken.Let the distance between $X$ and $Y$ be $d$.Time taken to

Vedclass · Accepted Answer

$\frac{2}{\bar v} = \frac{1}{v_1} + \frac{1}{v_2}$

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(C) Average speed is defined as the ratio of total distance traveled to the total time taken.Let the distance between $X$ and $Y$ be $d$.Time taken to travel from $X$ to $Y$ is $t_1 = \frac{d}{v_1}$.Time taken to travel from $Y$ to $X$ is $t_2 = \frac{d}{v_2}$.Total distance $= d + d = 2d$.Total time $= t_1 + t_2 = \frac{d}{v_1} + \frac{d}{v_2}$.Average speed $\bar v = \frac{	ext{Total distance}}{	ext{Total time}} = \frac{2d}{\frac{d}{v_1} + \frac{d}{v_2}}$.Canceling $d$ from the numerator and denominator,we get $\bar v = \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}}$.Rearranging this gives $\frac{2}{\bar v} = \frac{1}{v_1} + \frac{1}{v_2}$.

$A$ car moves from $X$ to $Y$ with a uniform speed $v_1$ and returns from $Y$ to $X$ with a uniform speed $v_2$. The average speed for this round trip is

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