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A biker travels $\frac{1}{3}$ of the distance $L$ with speed $v_1$ and $\frac{2}{3}$ of the distance with speed $v_2$. Then the average speed is
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Verified Answer
The correct answer is:
$\frac{3 v_1 v_2}{2 v_1+v_2}$
The given situation is shown below
Average speed $=\frac{\text { Total distance }}{\text { Time taken }}$
In given situation time to cover first part,
$t_1=\frac{L / 3}{v_1}=\frac{L}{3 v_1}$ $(\therefore$ Time $=$ Distance $/$ Speed $)$
and time to cover second part of journey,
$t_2=\frac{2 L / 3}{v_2}=\frac{2 L}{3 v_2}$
So, average speed for complete journey
$=\left(\frac{L / 3+2 L / 3}{\frac{L}{3 v_1}+\frac{2 L / 3}{v_2}}\right)=\frac{3 v_1 v_2}{2 v_1+v_2}$
Average speed $=\frac{\text { Total distance }}{\text { Time taken }}$
In given situation time to cover first part,
$t_1=\frac{L / 3}{v_1}=\frac{L}{3 v_1}$ $(\therefore$ Time $=$ Distance $/$ Speed $)$
and time to cover second part of journey,
$t_2=\frac{2 L / 3}{v_2}=\frac{2 L}{3 v_2}$
So, average speed for complete journey
$=\left(\frac{L / 3+2 L / 3}{\frac{L}{3 v_1}+\frac{2 L / 3}{v_2}}\right)=\frac{3 v_1 v_2}{2 v_1+v_2}$
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