A person travels x distance with velocity v_1 | vedclass

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(C) The average velocity is defined as the total displacement divided by the total time taken.Total distance $= x + x = 2x$.Time taken for the first p

Vedclass · Accepted Answer

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

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(C) The average velocity is defined as the total displacement divided by the total time taken.Total distance $= x + x = 2x$.Time taken for the first part of the journey $t_1 = \frac{x}{v_1}$.Time taken for the second part of the journey $t_2 = \frac{x}{v_2}$.Total time taken $T = t_1 + t_2 = \frac{x}{v_1} + \frac{x}{v_2} = x \left( \frac{1}{v_1} + \frac{1}{v_2} ight)$.Average velocity $v = \frac{	ext{Total distance}}{	ext{Total time}} = \frac{2x}{x \left( \frac{1}{v_1} + \frac{1}{v_2} ight)}$.Canceling $x$ from the numerator and denominator,we get $v = \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}}$.Rearranging this gives $\frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2}$.

$A$ person travels $x$ distance with velocity $v_1$ and then $x$ distance with velocity $v_2$ in the same direction. The average velocity of the person is $v$. Then the relation between $v, v_1$ and $v_2$ will be:

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