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

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(C) The average speed is defined as the total distance divided by the total time taken.Let the total distance be $S$.The distance for each segment is

Vedclass · Accepted Answer

$\frac{18}{11}V$

Vedclass · Answer

(C) The average speed is defined as the total distance divided by the total time taken.Let the total distance be $S$.The distance for each segment is $S/3$.The time taken for the first segment is $t_1 = \frac{S/3}{V} = \frac{S}{3V}$.The time taken for the second segment is $t_2 = \frac{S/3}{2V} = \frac{S}{6V}$.The time taken for the third segment is $t_3 = \frac{S/3}{3V} = \frac{S}{9V}$.Total time $T = t_1 + t_2 + t_3 = \frac{S}{3V} + \frac{S}{6V} + \frac{S}{9V}$.Taking the common denominator as $18V$,we get $T = \frac{6S + 3S + 2S}{18V} = \frac{11S}{18V}$.Average speed $V_{av} = \frac{S}{T} = \frac{S}{11S / 18V} = \frac{18V}{11}$.

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

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