A particle covers half of its total distance | vedclass

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Question

(C) Let the total distance be $d$. The particle covers the first half distance $(d/2)$ with speed $v_1$ and the second half distance $(d/2)$ with spee

Vedclass · Accepted Answer

$\frac{2 v_1 v_2}{v_1 + v_2}$

Vedclass · Answer

(C) Let the total distance be $d$. The particle covers the first half distance $(d/2)$ with speed $v_1$ and the second half distance $(d/2)$ with speed $v_2$.Time taken for the first half,$t_1 = \frac{d/2}{v_1} = \frac{d}{2v_1}$.Time taken for the second half,$t_2 = \frac{d/2}{v_2} = \frac{d}{2v_2}$.Average speed is defined as the total distance divided by the total time taken:$v_{av} = \frac{	ext{Total Distance}}{	ext{Total Time}} = \frac{d}{t_1 + t_2}$.Substituting the values of $t_1$ and $t_2$:$v_{av} = \frac{d}{\frac{d}{2v_1} + \frac{d}{2v_2}} = \frac{d}{\frac{d}{2} (\frac{1}{v_1} + \frac{1}{v_2})} = \frac{1}{\frac{1}{2} (\frac{v_1 + v_2}{v_1 v_2})}$.$v_{av} = \frac{2 v_1 v_2}{v_1 + v_2}$.

$A$ particle covers half of its total distance with speed $v_1$ and the rest half distance with speed $v_2$. Its average speed during the complete journey is

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