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Homework Help: Quadratic Word Problem

  1. Nov 4, 2006 #1
    Can you please help me with this quadratic word problem?

    An airplane flies 520 km against the win and 680 kim with the wind in a total time of 4hr. The speed of the airplane in still air is 300kim/hr. What is the speed of the wind?

    Any suggestion is greatly appreciated!!
  2. jcsd
  3. Nov 4, 2006 #2


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    We require demonstration of some effort on your part.

    Can you tell me how the speed of the airplane changes traveling with or against the wind?

    Can you write expressions that shows the 2 speeds.

    What is the relationship between speed, time and distance traveled?
    Last edited: Nov 4, 2006
  4. Nov 5, 2006 #3


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    Ok, since there is one unknown in the problem, i.e the speed of the wind. So if we let the speed of the wind be x (km / h)
    What's the speed of the airplane flying against the wind? (Note that the plane's speed in still air is 300 km/h).
    From there, how long does it take that plane to fly 520 km?
    What's the speed of the airplane flying with the wind?
    How long does it take that plane to fly 680 km?
    What is the total time that it takes?
    From there, can you find x?
    Can you go from here? :)
  5. Nov 5, 2006 #4
    Is this right?

    This is what I thought...

    made a chart with:

    DIstance (km) | Speed (km/hr) | Time (km)
    Against wind 520 300-x 520/300-x

    W/Wind 620 300+x 620/300+x

    Time=Distance/Speed (b/c Speed=distance/time)

    Let x km/hr be the speed of the wind.

    520/300-x + 620/300+x = 4

    figured out the equation and my answer was:
    Therefore, speed if wind is 55.7 km/hr

    Do you think this is right?
  6. Nov 5, 2006 #5


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    I found 2 possible wind speeds, 0 or 40 km/hr.

    You have the correct expression for the time of travel, except that you have changed the distance traveled with the wind to 620km, it was 680km in the initial problem statement.
  7. Nov 5, 2006 #6
    OMGSH I feel SO stupid! Thank you for noticing...I'm going to go back and change it..I'll let you know of my answer...thank you again!
  8. Nov 5, 2006 #7
    Ahhh..yes, I did get 0 & 40km...but obviously 0 does not work so therefore, 0 does not fit the question and the speed of the wind is 40 km/hr!
  9. Nov 5, 2006 #8
    or wait..could 0 fit the question...Im confused. We need to find the speed of the wind..Wind can't be zero because they're flying no?
  10. Nov 5, 2006 #9


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    But 0 DOES work. The individual leg times are different but add up to 4 hours. Just do the computation.
  11. Nov 6, 2006 #10


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    "Wind speed" is the speed of the wind relative to the ground. It has nothing to do with whether someone is flying or not.

    If the wind speed were 0 then the airplanes ground speed in both directions would be 300 km/h. it would take the airplane 520/300= 1.74 hours to fly 520 km and 680/300= 2.27 hours to fly 680 km: a total of 4 hours.
  12. Nov 6, 2006 #11
    When I plug 0 in to verify I get 4..but not when I plug 40 in..<4.5> is what I get.
  13. Nov 6, 2006 #12
    Does that mean 40 does not fit the question?
  14. Nov 6, 2006 #13
    Are you sure you plugged in 40 correctly?

    [tex]\frac{520}{300-x} + \frac{680}{300+x}[/tex]

    [tex]= \frac{520}{260} + \frac{680}{340}[/tex]

    [tex]= 2 + 2 = 4[/tex]

    Works for me.

    Since both solutions are valid, there are two solutions to the problem, as is frequently the case with a quadratic equation.
    Last edited: Nov 6, 2006
  15. Nov 6, 2006 #14
    I got 4!! yay!

    Thank you to everyone who has helped me:)
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