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27 July, 02:23

Matt and Anna Killian are frequent fliers on Fast-n-Go Airlines. They often fly between two cities that are a distance of 1575 miles apart. On one particular trip, they flew into the wind and the trip took 4.5 hours. The return trip with the wind behind them, only took about 3.5 hours. Find the speed of the wind and the speed of the plane in still air.

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  1. 27 July, 02:49
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    Speed of wind = 50mi/hr, Speed of plane in still air = 400mi/hr

    Explanation:

    Let the speed of the wind = Vw,

    Speed of the plane in still air = Vsa,

    The first trip the average speed of the plane = 1575mi/4.5hours = 350mi/hr

    The coming trip the wind behind = 1575mi/3.5hrs = 450

    Write the motion in equation form

    First trip (the plane flew into the wind)

    Vaverage = Vsa - Vw

    350 = Vsa - Vw

    Second trip the wind was behind

    450 = Vsa + Vw

    Adding the two equation

    800 = 2Vas

    Vas = 800/2 = 400mi/hr

    Substitute for Vas into equation 1

    350mi/hr = 400mi/hr - Vw

    Vw = 400-350 = 50mi/hr
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