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(D) Let the total distance be $D$.Distance covered with speed $v_1$ is $d_1 = 0.4D$.Distance covered with speed $v_2$ is $d_2 = 0.6D$.Time taken for t

Vedclass · Accepted Answer

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

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(D) Let the total distance be $D$.Distance covered with speed $v_1$ is $d_1 = 0.4D$.Distance covered with speed $v_2$ is $d_2 = 0.6D$.Time taken for the first part is $t_1 = \frac{d_1}{v_1} = \frac{0.4D}{v_1}$.Time taken for the second part is $t_2 = \frac{d_2}{v_2} = \frac{0.6D}{v_2}$.Average speed $v_{avg} = \frac{	ext{Total Distance}}{	ext{Total Time}} = \frac{D}{t_1 + t_2}$.$v_{avg} = \frac{D}{\frac{0.4D}{v_1} + \frac{0.6D}{v_2}} = \frac{1}{\frac{0.4}{v_1} + \frac{0.6}{v_2}}$.$v_{avg} = \frac{1}{\frac{0.4v_2 + 0.6v_1}{v_1 v_2}} = \frac{v_1 v_2}{0.4v_2 + 0.6v_1}$.Multiply numerator and denominator by $5$ to simplify: $v_{avg} = \frac{5 v_1 v_2}{2v_2 + 3v_1} = \frac{5 v_1 v_2}{3v_1 + 2v_2}$.

If a car travels $40 \%$ of the total distance with a speed $v_1$ and the remaining distance with a speed $v_2$,then the average speed of the car is:

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