A biker travels frac{1}{3} of the distance L | vedclass

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(B) The average speed is defined as the total distance divided by the total time taken.Let the total distance be $L$.The distance for the first part i

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$\frac{3 v_1 v_2}{2 v_1+v_2}$

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(B) The average speed is defined as the total distance divided by the total time taken.Let the total distance be $L$.The distance for the first part is $d_1 = \frac{L}{3}$ and the speed is $v_1$.The time taken for the first part is $t_1 = \frac{d_1}{v_1} = \frac{L/3}{v_1} = \frac{L}{3 v_1}$.The distance for the second part is $d_2 = \frac{2L}{3}$ and the speed is $v_2$.The time taken for the second part is $t_2 = \frac{d_2}{v_2} = \frac{2L/3}{v_2} = \frac{2L}{3 v_2}$.The total time taken is $T = t_1 + t_2 = \frac{L}{3 v_1} + \frac{2L}{3 v_2} = \frac{L}{3} \left( \frac{1}{v_1} + \frac{2}{v_2} ight) = \frac{L}{3} \left( \frac{v_2 + 2 v_1}{v_1 v_2} ight)$.The average speed $v_{avg} = \frac{	ext{Total distance}}{	ext{Total time}} = \frac{L}{\frac{L}{3} \left( \frac{2 v_1 + v_2}{v_1 v_2} ight)} = \frac{3 v_1 v_2}{2 v_1 + v_2}$.

$A$ 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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