A jetliner can fly 7.5 hours on a full load of fuel. Without any wind it flies at a speed of 2.14 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 70.1-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

the maximum distance (in kilometers) that the plane can travel due west and just be able to return home is; d_max = 2578km

Explanation:

We are told that the velocity of the plane is 2.14 x 10² m/s.

Let's convert it to km/h.

Thus;

Vp = 2.14 x 10²m/s x (1/1000)km/m x 3600s/h =

Vp = 770.14 km/h

Now,let's do the same for velocity of the wind;

Vw = 70.1 m/s x (1/1000)km/m x 3600s/h

Vw = 252.36 km/h

velocity of the plane on the departure flight so the velocity of the plane will be(770.14 - 252.36) = 517.78 km/h

on the way back or on return, the wind and plane velocity will add up. so the total velocity will be;

(770.14 + 252.36) = 1022.5 km/h

Now, we have been given total time as 7.5 hours and the distance both ways must be the same so:

distance there;d1 = 517.78t

distance back;d2 = 1022.5(7.5 - t)

So for d1 = d2, we have;

517.78t = (1022.5)(7.5 - t),

517.78t = 7668.75 - 1022.5t

Rearranging, we have;

517.78t + 1022.5t = 7668.75

1540.28t = 7668.75

t = 7668.75/1540.28

t = 4.979 hr

We know that;

Distance = velocity x time

Thus;

Maximum distance is;

d_max = Vt = 517.78 x 4.979

d_max = 2578km

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answer( sicar );