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(C) Let the total distance be $d$.Time taken for the first half distance $(d/2)$ is $t_1 = \frac{d/2}{V_1} = \frac{d}{2V_1}$.Time taken for the second

Vedclass · Accepted Answer

$\frac{2}{\frac{1}{V_1}+\frac{1}{V_2}}$

Vedclass · Answer

(C) Let the total distance be $d$.Time taken for the first half distance $(d/2)$ is $t_1 = \frac{d/2}{V_1} = \frac{d}{2V_1}$.Time taken for the second half distance $(d/2)$ is $t_2 = \frac{d/2}{V_2} = \frac{d}{2V_2}$.Average velocity is defined as the total distance divided by the total time taken.Average velocity $= \frac{	ext{Total distance}}{	ext{Total time}} = \frac{d}{t_1 + t_2}$.Substituting the values of $t_1$ and $t_2$:Average velocity $= \frac{d}{\frac{d}{2V_1} + \frac{d}{2V_2}} = \frac{d}{\frac{d}{2} \left( \frac{1}{V_1} + \frac{1}{V_2} ight)} = \frac{1}{\frac{1}{2} \left( \frac{V_1 + V_2}{V_1 V_2} ight)} = \frac{2V_1 V_2}{V_1 + V_2}$.Note that $\frac{2V_1 V_2}{V_1 + V_2}$ is equivalent to $\frac{2}{\frac{1}{V_1} + \frac{1}{V_2}}$,which matches option $C$.

If a person moving along a straight line path covers the first half distance with velocity $V_1$ and the next half distance with velocity $V_2$,then the average velocity of the person is

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