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(D) Let the total distance $AB$ be $d$.Each segment covers a distance of $\frac{d}{3}$.The time taken for each segment is $t_{1} = \frac{d/3}{v_{1}}$,

Vedclass · Accepted Answer

$18$

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(D) Let the total distance $AB$ be $d$.Each segment covers a distance of $\frac{d}{3}$.The time taken for each segment is $t_{1} = \frac{d/3}{v_{1}}$,$t_{2} = \frac{d/3}{v_{2}}$,and $t_{3} = \frac{d/3}{v_{3}}$.The average velocity $v_{avg}$ is given by:$v_{avg} = \frac{	ext{Total Distance}}{	ext{Total Time}} = \frac{d}{t_{1} + t_{2} + t_{3}} = \frac{d}{\frac{d}{3v_{1}} + \frac{d}{3v_{2}} + \frac{d}{3v_{3}}}$$v_{avg} = \frac{3}{\frac{1}{v_{1}} + \frac{1}{v_{2}} + \frac{1}{v_{3}}}$Given $v_{1} = 11 \, ms^{-1}$,$v_{2} = 2v_{1} = 22 \, ms^{-1}$,and $v_{3} = 3v_{1} = 33 \, ms^{-1}$.Substituting these values:$v_{avg} = \frac{3}{\frac{1}{11} + \frac{1}{22} + \frac{1}{33}} = \frac{3}{\frac{6+3+2}{66}} = \frac{3 	imes 66}{11} = 3 	imes 6 = 18 \, ms^{-1}$.

$A$ car covers distance $AB$ with the first one-third at velocity $v_{1} \, ms^{-1}$,the second one-third at $v_{2} \, ms^{-1}$,and the last one-third at $v_{3} \, ms^{-1}$. If $v_{3}=3v_{1}$,$v_{2}=2v_{1}$,and $v_{1}=11 \, ms^{-1}$,then the average velocity of the car is . . . . . . $ms^{-1}$.

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