Given, the speed in the first trip, S1=40 miles/h.
The speed in the return trip, S2=50 miles/h.
The distance is the same for both trips.
When the distance is same, the average speed of the drive is,
[tex]\begin{gathered} S=\frac{2S1S2}{S1+S2} \\ =\frac{2\times40\times50}{40+50} \\ =\frac{4000}{90} \\ =\frac{400}{9}\text{ miles/h} \end{gathered}[/tex]Given, the time taken for entire trip is t=9 h.
Now, the distance travelled both ways to park is,
[tex]\begin{gathered} D=St \\ =\frac{400}{9}\times9 \\ =400\text{ miles} \end{gathered}[/tex]Hence, the distance travelled one way to the park is,
[tex]\begin{gathered} d=\frac{D}{2} \\ =\frac{400}{2} \\ =200\text{ miles} \end{gathered}[/tex]Therefore, the distance travelled one way to the park is 200 miles.
