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Related rates and flying airplane

  1. May 15, 2009 #1
    1. The problem statement, all variables and given/known data
    An airplane flying horizontally at an altitude of 3 miles and at a speed of 480mi/hr passes directly above an observer on the ground. How fast is the distance from the observer to the airplane increasing 30 seconds later?



    2. Relevant equations



    3. The attempt at a solution
    can you check if this is right:
    If the plane is moving at 480 mi/hr, use 30 seconds to compute how far it traveled. (480 miles/hour)(hour/60 minutes) = 8 miles/minute.
    It can be seen that 30 seconds is half a minutes, so the plane would have flown half of 8 miles. You can probably do this, but 8/2=4.
    If the plane were 3 miles overhead and travelled 4 miles away, the distance can be found as √(3²+4²). Compute 3². Compute 4². Add the two together. The number should be recognized as X².


    rest i dont know
     
  2. jcsd
  3. May 15, 2009 #2

    Dick

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    Homework Helper

    It's a rates problem. The vertical distance is 3mi. The horizontal distance is h(t) where h depends on time because the plane is moving. Right? They want to know what is the rate of change of sqrt((3mi)^2+h(t)^2). Take a derivative. Can you write an expression for h(t)?
     
  4. May 16, 2009 #3
    hi !

    yr you can derive a equation for the distance between observer and airplane

    it will be like this
    D=sqrt(3^2+(4T/30)^2)
    hope you will know how to approach it ,here T means the time in seconds.

    then just differentiate D w.r.t T
    then u will get dD/dT
    just subb T in the above derivative and u will get the value for the change of distance with time .i think that what u r looking for ...........

    it is great to help a sri lankan!!reply if u want more help
     
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