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,
[tex]v_{ab}=5\text{ m/s}[/tex]The velocity of man going from point B to point A is,
[tex]v_{ba}=4\text{ m/s}[/tex]The time is given as,
[tex]t=\frac{\text{ distance}}{\text{ speed}}[/tex]Therefore, time taken by man going from point A to point B is given as,
[tex]t_{ab}=\frac{d}{v_{ab}}[/tex]The time taken by man going from point B to point A is given as,
[tex]t_{ba}=\frac{d}{v_{ba}}[/tex]The average speed is given as,
[tex]\begin{gathered} v_{avg}=\frac{\text{ total distance covered}}{\text{ total time taken}} \\ =\frac{d+d}{t_{ab}+t_{ba}} \\ =\frac{2d}{\frac{d}{v_{ab}}+\frac{d}{v_{ba}}} \\ =\frac{2d}{d(\frac{1}{v_{ab}}+\frac{1}{v_{ba}})} \end{gathered}[/tex]Simplifying the above equation;
[tex]\begin{gathered} v_{avg}=\frac{2}{\frac{1}{v_{ab}}+\frac{1}{v_{ba}}} \\ =\frac{2}{\frac{v_{ba}+v_{ab}}{v_{ab}v_{ba}}} \\ =\frac{2v_{ab}v_{ba}}{v_{ab}+v_{ba}} \end{gathered}[/tex]Substituting all known values,
[tex]\begin{gathered} v_{avg}=\frac{2\times(5\text{ m/s})\times(4\text{ m/s})}{(5\text{ m/s})+(4\text{ m/s})} \\ =4.44\text{ m/s} \end{gathered}[/tex]Therefore, the average speed of the man is 4.44 m/s.
