A biker travels $\frac{1}{3}$ of the distance $L$ with speed $v_1$ and $\frac{2}{3}$ of the distance with…

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
  1. $\frac{v_1 v_2}{v_1+v_2}$
  2. $\frac{3 v_1 v_2}{2 v_1+v_2}$
  3. $\frac{3 v_1 v_2}{v_1+2 v_2}$
  4. $\frac{v_1+v_2}{v_1 v_2}$

Solution

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 $=\left(\frac{L / 3+2 L / 3}{\frac{L}{3 v_1}+\frac{2 L / 3}{v_2}}\right)=\frac{3 v_1 v_2}{2 v_1+v_2}$

Asked in: AP EAMCET 2022 (07 Jul Shift 1)

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