A body moves in a straight line with speed $v_1$ and $v_2$ for distance which are in ratio $1: 2$. Find the average speed.

• A

  $3 v_1+v_2$
• B

  $v_2+2 v_1$
• Correct answer
  C

  $\frac{3 v_1 v_2}{v_2+2 v_1}$
• D

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

Solution

Given that, the distances travelled by the body are in ratio $1: 2$. Let us suppose that $s$ is the distance, the body travels with speed $v_1$. Then, $2 s$ will be the distance travelled by the body with speed $v_2$. This scenario is shown in the line-diagram below
For $A$ to $C$, $\begin{aligned} & \text { Average speed }=\frac{\text { Total distance travelled }}{\text { Total time taken }} \\ & \qquad \begin{aligned} v_{\text {avg }} & =\frac{s_1+s_2}{t_1+t_2}=\frac{s+2 s}{\frac{s}{v_1}+\frac{2 s}{v_2}} \\ & =\frac{3 s}{s\left[\frac{1}{v_1}+\frac{2}{v_2}\right]}=\frac{3}{\left[\frac{v_2+2 v_1}{v_1 v_2}\right]}=\frac{3 v_1 v_2}{v_2+2 v_1} \end{aligned} \end{aligned}$

Hence, the correct answer is option: 'C'.