Question

A man goes around a field at an average rate of 5.0 m/s, and a second time at an average rate of 4.0m/s. What is his average speed in the two round trip?

Answer 1

Let A and B are the starting and end point of field respectively.

The velocity of man going from point A to point B is,

`v_(ab)=5\text{ m/s}`

The velocity of man going from point B to point A is,

`v_(ba)=4\text{ m/s}`

The time is given as,

`t=\frac{\text{ distance}}{\text{ speed}}`

Therefore, time taken by man going from point A to point B is given as,

`t_(ab)=(d)/(v_(ab))`

The time taken by man going from point B to point A is given as,

`t_(ba)=(d)/(v_(ba))`

The average speed is given as,

`\begin{gathered} v_(avg)=\frac{\text{ total distance covered}}{\text{ total time taken}} \\ =(d+d)/(t_(ab)+t_(ba)) \\ =(2d)/((d)/(v_(ab))+(d)/(v_(ba))) \\ =(2d)/(d((1)/(v_(ab))+(1)/(v_(ba)))) \end{gathered}`

Simplifying the above equation;

`\begin{gathered} v_(avg)=(2)/((1)/(v_(ab))+(1)/(v_(ba))) \\ =(2)/((v_(ba)+v_(ab))/(v_(ab)v_(ba))) \\ =(2v_(ab)v_(ba))/(v_(ab)+v_(ba)) \end{gathered}`

Substituting all known values,

`\begin{gathered} v_(avg)=\frac{2*(5\text{ m/s})*(4\text{ m/s})}{(5\text{ m/s})+(4\text{ m/s})} \\ =4.44\text{ m/s} \end{gathered}`

Therefore, the average speed of the man is 4.44 m/s.