A body covers the first one-third of the dist | vedclass

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(A) Let the total distance be $3d$. The distance for each part is $d$.The time taken for the first part is $t_1 = \frac{d}{20}$.The time taken for the

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$28$

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(A) Let the total distance be $3d$. The distance for each part is $d$.The time taken for the first part is $t_1 = \frac{d}{20}$.The time taken for the second part is $t_2 = \frac{d}{30}$.The time taken for the third part is $t_3 = \frac{d}{40}$.The average velocity $v_{avg}$ is given by the total distance divided by the total time:$v_{avg} = \frac{3d}{t_1 + t_2 + t_3} = \frac{3d}{\frac{d}{20} + \frac{d}{30} + \frac{d}{40}}$$v_{avg} = \frac{3}{\frac{1}{20} + \frac{1}{30} + \frac{1}{40}}$Taking the least common multiple of $20, 30, 40$,which is $120$:$v_{avg} = \frac{3}{\frac{6 + 4 + 3}{120}} = \frac{3 	imes 120}{13} = \frac{360}{13}$$v_{avg} \approx 27.69 \, m/s \approx 28 \, m/s$.

$A$ body covers the first one-third of the distance with a velocity of $20 \, m/s$,the second one-third with a velocity of $30 \, m/s$,and the last one-third with a velocity of $40 \, m/s$. The average velocity is nearly........$m/s$.

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