A body has speed V, 2V and 3V in the first 1/ | vedclass

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Question

(C) The total distance is $S$. The distance is divided into three equal parts,each of length $S/3$.Let the speeds in these parts be $v_1 = V$,$v_2 = 2

Vedclass · Accepted Answer

$\frac{18}{11}V$

Vedclass · Answer

(C) The total distance is $S$. The distance is divided into three equal parts,each of length $S/3$.Let the speeds in these parts be $v_1 = V$,$v_2 = 2V$,and $v_3 = 3V$.The time taken for each part is $t_1 = \frac{S/3}{V}$,$t_2 = \frac{S/3}{2V}$,and $t_3 = \frac{S/3}{3V}$.Average speed $V_{av} = \frac{	ext{Total Distance}}{	ext{Total Time}} = \frac{S}{t_1 + t_2 + t_3}$.$V_{av} = \frac{S}{\frac{S}{3V} + \frac{S}{6V} + \frac{S}{9V}} = \frac{1}{\frac{1}{3V} + \frac{1}{6V} + \frac{1}{9V}}$.Taking the common denominator $18V$: $V_{av} = \frac{1}{\frac{6+3+2}{18V}} = \frac{18V}{11}$.

$A$ body has speed $V, 2V$ and $3V$ in the first $1/3$ of distance $S$,second $1/3$ of $S$,and third $1/3$ of $S$ respectively. Its average speed will be:

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