To solve this we are going to use the speed equation: [tex]S= \frac{d}{t} [/tex]
where
[tex]S[/tex] is speed
[tex]d[/tex] is distance
[tex]t[/tex] is time
First, we are going to label the information that the problem is giving us:
Since the plane has enough fuel for a five our flight, the total time of the round trip will be five hours; therefore, [tex]t_{t}=5[/tex]. We know that the speed of the plane with the wind is 225 miles per hour, so [tex]S_{ww}=225[/tex]. We also know that the speed of the plane against the wind is 180 miles per hour, so [tex]S_{aw}=180[/tex]. Notice that Since we are dealing with a round trip, the distance against the wind and the distance with the wind will be the same.
For the trip with the wind:
[tex]S_{ww}= \frac{d}{t_{w}} [/tex]
[tex]225= \frac{d}{t_{w}} [/tex] equation (1)
For the trip against the wind:
[tex]S_{aw}= \frac{d}{t_{a} } [/tex]
[tex]180= \frac{d}{t_{a} } [/tex] equation (2)
Since the whole trip takes 5 hours, [tex]t_{w}+t_{a}=5[/tex]
Solving for [tex]t_{w}[/tex]:
[tex]t_{w}=5-t_{a}[/tex] equation (3)
Replacing (3) in (1):
[tex]225= \frac{d}{t_{w}} [/tex]
[tex]225= \frac{d}{5-t_{a}} [/tex] equation (4)
Solving for [tex]d[/tex] in (2)
[tex]180= \frac{d}{t_{a} } [/tex]
[tex]d=180t_{a}[/tex] equation (5)
Replacing (5) in (4):
[tex]225= \frac{d}{5-t_{a}} [/tex]
[tex]225= \frac{180t_{a}}{5-t_{a}} [/tex]
[tex]225(5-t_{a})=180t_{a}[/tex]
[tex]1125-225t_{a}=180t_{a}[/tex]
[tex]405t_{a}=1125[/tex]
[tex]t_{a}= \frac{1125}{405} [/tex]
[tex]t_{a}= \frac{25}{9} [/tex] equation (6)
Replacing (6) in (5):
[tex]d=180t_{a}[/tex]
[tex]d=180( \frac{25}{9} )[/tex]
[tex]d=500[/tex] miles
We can conclude that the plane can travel a distance of 500 miles to make a round trip if it has just enough fuel for a five hour flight.
