A jetliner can fly 8.0 hours on a full load of fuel. Without any wind it flies at a speed of 2.42 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 40.7-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

3386.23 Km

Explanation:

Without any wind interference the speed of the plane, it flies at 2.42 x 102 m/s or 242 m/s

The magnitude of the velocity of the air = 40.7-m/s

The time taken, for the plane to fly due west, tw = distance, x ÷ (speed of the plane without wind interference, up - the magnitude of the velocity of the air, ua) (equation 1)

The time taken, for the plane to fly due east, te = distance, x ÷ (speed of the plane without wind interference, up + the magnitude of the velocity of the air, ua) (equation 2)

To calculate the total time, t, the plane can fly to cover east and west distances, we add equation 1 and 2 together

t = (x ÷ (up - ua)) + (x ÷ (up + ua))

Making distance x, the subject of the formula we have:

x = t ( up²- ua²) ÷ 2up

note t = 8.0 hours which is (8 x 60 x 60) seconds = 28,800‬

So, x = 28,800‬ ((242 m/s)² - (40.7-m/s)²)÷ 2(242 m/s)

x= 1,638,936,288‬ ÷ 484

= 3386.23 Km

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