A jetliner can fly 5.6 hours on a full load of fuel. Without any wind it flies at a speed of 2.25 x 102 m/s. The plane is to make a round-trip by heading due west for a certain distance, turning around, and then heading due east for the return trip. During the entire flight, however, the plane encounters a 71.9-m/s wind from the jet stream, which blows from west to east. What is the maximum distance (in kilometers) that the plane can travel due west and just be able to return home?

If [tex]v_P=2.25\times10^2m/s[/tex] is the velocity of the plane without any wind and [tex]v_W=71.9m/s[/tex] is the velocity of the wind (which blows from west to east), then the velocity of the plane relative to the ground when it goes from west to east is [tex]v_{WE}=v_P+v_W[/tex] and when it goes from east to west is [tex]v_{EW}=v_P-v_W[/tex].

We know that the distance traveled from west to east must be the same as from east to west, so we have:

[tex]d_{EW}=d_{WE}[/tex]

Which means (since v=d/t):

[tex]v_{EW}t_{EW}=v_{WE}t_{WE}[/tex]

The time from west to east plust the time from east to west must be the total time t (which we will set as the maximum time the plane can fly), so we can write [tex]t_{WE}=t-t_{EW}[/tex], and substituting:

[tex]v_{EW}t_{EW}=v_{WE}(t-t_{EW})[/tex]

Which means:

[tex]v_{EW}t_{EW}=v_{WE}t-v_{WE}t_{EW}[/tex]

[tex]v_{WE}t_{EW}+v_{EW}t_{EW}=v_{WE}t[/tex]

[tex](v_{WE}+v_{EW})t_{EW}=v_{WE}t[/tex]

[tex]t_{EW}=\frac{v_{WE}t}{v_{WE}+v_{EW}}[/tex]

We now substitute what we had from the beginning:

[tex]t_{EW}=\frac{(v_P+v_W)t}{v_P+v_W+v_P-v_W}}=\frac{(v_P+v_W)t}{2v_P}}[/tex]

Using our values, and considering that 5.6 hours are 20160 seconds:

[tex]t_{EW}=\frac{(2.25\times10^2m/s+71.9m/s)(20160s)}{2(2.25\times10^2m/s)}}=13301s[/tex]

Finally, the distance due west will be:

[tex]d_{EW}=v_{EW}t_{EW}=(v_P-v_W)t_{EW}=(2.25\times10^2m/s-71.9m/s)(13301s)=2036383.1m=2036Km[/tex]