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Math Help - Algebra - Wind problem

  1. #1
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    Algebra - Wind problem

    I can't figure this problem out.

    During a 3000 mile flight, a plane encountered a strong tail wind that increased its speed by 100 mi/hour. This increase of speed shortened the flying time by one hour. Find the speed of the plane relative to the ground.
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  2. #2
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    Hello, SwissArmelle!

    I'll do it the "standard" way", but we'll get a tricky equation . . .


    During a 3000-mile flight, a plane encountered a strong tail wind
    that increased its speed by 100 mi/hour.
    This increase of speed shortened the flying time by one hour.
    Find the speed of the plane relative to the ground.
    We will use: . \text{Distance} \:=\: \text{Speed} \times \text{Time}\quad\Rightarrow\quad t \:=\:\frac{d}{s}

    Let S = speed of the plane (ground speed).

    Normally, it would fly the 3000 miles at S mph.
    . . It would take: . \frac{3000}{S} hours.

    With the tailwind, its speed is S + 100 mph.
    . . To fly 3000 miles, it took only: . \frac{3000}{S + 100} hours.
    And this is one hour less than its normal time.

    There is our equation . . . . \frac{3000}{S + 100} \:=\:\frac{3000}{S} - 1


    Multiply by S(S+100)\!:\;\;3000S \:=\:3000(S + 100) -S(S + 100)

    This simplifies to the quadratic: . S^2 + 100S - 30,000 \:=\:0

    . . which factors: . (S - 500)(S + 600) \:=\:0

    . . and has roots: . S \:=\:500,\:-600


    Therefore, the plane's ground speed is: \boxed{500\:mph}

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