A car moving on a straight road covers one-th | vedclass

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(D) Let the total distance be $x$.Time taken to cover the first one-third distance $(x/3)$ at $v_1 = 20 \, km/hr$ is $t_1 = \frac{x/3}{20} = \frac{x}{

Vedclass · Accepted Answer

$36$

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(D) Let the total distance be $x$.Time taken to cover the first one-third distance $(x/3)$ at $v_1 = 20 \, km/hr$ is $t_1 = \frac{x/3}{20} = \frac{x}{60} \, hr$.Time taken to cover the remaining two-thirds distance $(2x/3)$ at $v_2 = 60 \, km/hr$ is $t_2 = \frac{2x/3}{60} = \frac{2x}{180} = \frac{x}{90} \, hr$.Average speed is defined as the total distance divided by the total time: $v_{avg} = \frac{x}{t_1 + t_2}$.$v_{avg} = \frac{x}{\frac{x}{60} + \frac{x}{90}} = \frac{1}{\frac{1}{60} + \frac{1}{90}}$.Finding a common denominator: $\frac{1}{60} + \frac{1}{90} = \frac{3+2}{180} = \frac{5}{180} = \frac{1}{36}$.Therefore,$v_{avg} = \frac{1}{1/36} = 36 \, km/hr$.

$A$ car moving on a straight road covers one-third of the distance with $20 \, km/hr$ and the rest with $60 \, km/hr$. The average speed is .......... $km/hr$.

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