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19 October, 16:14

Dave and Sandy Hartranft are frequent flyers with a particular airline. They often fly from City A to City B, a distance of 756 miles. On one particular trip, they fly into the wind, and the the flight takes 2 hours. The return trip, with the wind behind them, only takes1 1/2 hours. If the wind speed is the same on each trip, find the speed of the wind and find the speed of the plane in still air.

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  1. 19 October, 16:21
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    Plane = 441 mi/h; wind = 63 mi/h

    Step-by-step explanation:

    distance = rate * time

    Let p = the plane's speed in still air

    and w = the wind speed. Then,

    p + w = speed with wind

    p - w = speed against wind

    We have two conditions:

    (1) 756 = (p - w) * 2

    (2) 756 = (p + w) * 1.5

    Distribute the constants (3) 756 = 2p - 2w

    (4) 756 = 1.5p + 1.5 w

    Multiply Equation (3) by 3 (5) 2268 = 6p - 6w

    Multiply Equation (4) by 4 (6) 3024 = 6p + 6w

    Add Equations (5) and (6) 5292 = 12p

    Divide each side by 12 (7) p = 441 mi/h

    Substitute (7) into Equation (3) 756 = 882 - 2w

    Add 2w to each side 2w + 756 = 882

    Subtract 756 from each side 2w = 126

    Divide each side by 2 w = 63 mph

    The plane's speed in still air is 441 mi/h.

    The wind speed is 63 mph.

    Check:

    (1) 756 = (441 - 63) * 2 (2) 756 = (441 + 63) * 1.5

    756 = 378 * 2 756 = 504 * 1.5

    756 = 756 756 = 756
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