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11 October, 06:07

An airplane took 4 hours to fly 2400 miles against headwind. The return trip with the wind took 3 hours. Find the speed of the plane in still air and the speed of the wind.

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  1. 11 October, 06:14
    0
    Speed in still air: x

    speed of wind: y

    speed against headwind is x-y = 2400/4=600

    speed of return trip x+y = 2400/3=800

    solve the linear equation:

    add the two equations to eliminate y:

    2x=1400

    x=700

    y=100
  2. 11 October, 06:16
    0
    Use distance = rate x time

    For the trip there:

    2400 = rate x 4

    where rate = A (airplane speed) - W (wind speed)

    so:

    2400 = (A - W) x 4

    The return trip:

    2400 = (A + W) x 3

    Now solve. Since they both equal 2400, I'll set them equal to each other:

    (A - W) x 4 = (A + W) x 3

    4A - 4W = 3A + 3W

    A = 7W

    Now substitute back into one of the original equations:

    2400 = (7W - W) x 4

    2400 = 6W x 4

    2400 = 24W

    100 = W

    Therefore A = 700

    So airplane speed in still air is 700 mph and the wind speed is 100 mph.
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