If a car covers 2/5^{th} of the total distanc | vedclass

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(D) Average speed is defined as the total distance traveled divided by the total time taken.Let the total distance be $x$.The time taken to cover the

Vedclass · Accepted Answer

$\frac{5 v_1 v_2}{3 v_1 + 2 v_2}$

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(D) Average speed is defined as the total distance traveled divided by the total time taken.Let the total distance be $x$.The time taken to cover the first part ($2/5$ of $x$) is $t_1 = \frac{2x/5}{v_1} = \frac{2x}{5v_1}$.The time taken to cover the second part ($3/5$ of $x$) is $t_2 = \frac{3x/5}{v_2} = \frac{3x}{5v_2}$.Total time $T = t_1 + t_2 = \frac{2x}{5v_1} + \frac{3x}{5v_2} = \frac{x}{5} \left( \frac{2}{v_1} + \frac{3}{v_2} ight) = \frac{x(2v_2 + 3v_1)}{5v_1 v_2}$.Average speed $v_{avg} = \frac{	ext{Total distance}}{	ext{Total time}} = \frac{x}{\frac{x(2v_2 + 3v_1)}{5v_1 v_2}} = \frac{5v_1 v_2}{3v_1 + 2v_2}$.

If a car covers $2/5^{th}$ of the total distance with speed $v_1$ and $3/5^{th}$ of the total distance with speed $v_2$,then the average speed is:

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