Question

A person first walks at a constant speed $v_1$ along a straight line from $A$ to $B$, and then back along the line from $B$ to $A$ with constant speed $v_2$. What is the average speed for the entire trip?

Answer 1

Step 1: Given Data

1. $v_1$ is the speed from point $A$ to point $B$
2. $v_2$ is the speed from point $B$ to point $A$

Step 2: Find the average speed ${\mathit{v}}_{\mathbf{a}\mathbf{v}}$ for the entire trip

Assume the distance between point $A$ and point $B$ as the total distance from the point $A$ to point $B$ and point $B$ to point $A$$=2s$

As we know, $time=\frac{dis\tan ce}{speed}$

So, $t_1$ $=\frac{s}{v_1}$ and $t_2$ $=\frac{s}{v_2}$

Here, Time $t_1$ = time taken from point $A$ to point $B$

Time $t_2$ = time taken from point $B$ to point $A$

and $averagespeed=\frac{Totaldis\tan ce}{Totaltime}$

$$\begin{gathered} \Rightarrow v_{av}=\frac{2s}{{\displaystyle \frac{s}{v_1}}+{\displaystyle \frac{s}{v_2}}} \\ =\frac{2v_1{v}_2}{v_1+v_2} \end{gathered}$$

Final Answer:

The average speed is $\frac{\mathbf{2}{\mathbf{v}}_{\mathbf{1}}{\mathbf{v}}_{\mathbf{2}}}{{\mathbf{v}}_{\mathbf{1}}\mathbf{+}{\mathbf{v}}_{\mathbf{2}}}$ for the entire trip.