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Thread: 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: .$\displaystyle \text{Distance} \:=\: \text{Speed} \times \text{Time}\quad\Rightarrow\quad t \:=\:\frac{d}{s}$

    Let $\displaystyle S$ = speed of the plane (ground speed).

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

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

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


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

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

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

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


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

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