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An airliner flies against the wind on one leg of its round trip in 5.5 hours. It flies back to its starting​ point, with the wind in 5 hours. If the average speed of the wind is 21 miles per​ hour, what is the speed of the airliner in still​ air?

1 Answer

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let's call the speed of the airliner in still air "a" in miles per hour. when flying against the wind, its effective speed is reduced, so it travels at (a - 21) miles per hour. when flying with the wind, its effective speed is increased, so it travels at (a + 21) miles per hour.

we can use the formula: time = distance / speed.

for the first leg of the trip (against the wind), the time is 5.5 hours, and for the return leg (with the wind), the time is 5 hours. since the distance is the same for both legs of the trip, we can set up the following equations:

1. distance = (a - 21) * 5.5
2. distance = (a + 21) * 5

since both equations represent the same distance, we can set them equal to each other:

(a - 21) * 5.5 = (a + 21) * 5

now, let's solve for "a":

5.5a - 115.5 = 5a + 105

subtract 5a from both sides:

0.5a - 115.5 = 105

add 115.5 to both sides:

0.5a = 220.5

now, divide by 0.5 to find "a":

a = 441

so, the speed of the airliner in still air is 441 miles per hour.

hope this helped!
User BurningLights
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