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(D) Let the total distance be $x$.The distance covered in each part is $x/3$.The time taken for the first part is $t_1 = \frac{x/3}{v_1} = \frac{x}{3v

Vedclass · Accepted Answer

$\frac{3 v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$

Vedclass · Answer

(D) Let the total distance be $x$.The distance covered in each part is $x/3$.The time taken for the first part is $t_1 = \frac{x/3}{v_1} = \frac{x}{3v_1}$.The time taken for the second part is $t_2 = \frac{x/3}{v_2} = \frac{x}{3v_2}$.The time taken for the third part is $t_3 = \frac{x/3}{v_3} = \frac{x}{3v_3}$.Average speed $v_{av} = \frac{	ext{Total distance}}{	ext{Total time}} = \frac{x}{t_1 + t_2 + t_3}$.$v_{av} = \frac{x}{\frac{x}{3v_1} + \frac{x}{3v_2} + \frac{x}{3v_3}} = \frac{1}{\frac{1}{3v_1} + \frac{1}{3v_2} + \frac{1}{3v_3}}$.$v_{av} = \frac{3}{\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}} = \frac{3}{\frac{v_2 v_3 + v_1 v_3 + v_1 v_2}{v_1 v_2 v_3}}$.$v_{av} = \frac{3 v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$.

If a body covers one-third of its total distance at speed $v_1$,the next one-third at speed $v_2$,and the last one-third at speed $v_3$,then the average speed will be:

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