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It's in my Exam|Optimization Question|Picture Included

  1. Mar 22, 2007 #1
    1. The problem statement, all variables and given/known data
    Ship B is going west at 12km\h[W] while Ship A is going south at 9km\h. In the beginning, the distance that they are away from each other is 75km. When will the ships be the closest?

    2. The attempt at a solution






    And t can't be a negative number.
    Last edited: Mar 23, 2007
  2. jcsd
  3. Mar 22, 2007 #2
    Your diagram is wrong. You have ship A going south and ship B going west. Anywho, to solve this problem, determine an equation that will give you the distance D between ship A and ship B as observed from one of the ships.
  4. Mar 22, 2007 #3


    then what?
    Last edited: Mar 23, 2007
  5. Mar 22, 2007 #4
    How are you deriving those equations?
  6. Mar 22, 2007 #5
    This equation was better except for an error in the first term on the RHS.
  7. Mar 22, 2007 #6
    I saw those equation used in a textbook in a similar question.

    What's RHS?

    And what do I now after setting up the equation for D?



    Is this what I do?
    Last edited: Mar 23, 2007
  8. Mar 23, 2007 #7
    It means right hand side.

    You have not explained how you derived that equation.
  9. Mar 23, 2007 #8
    It's the pythagorean theorem.
  10. Mar 23, 2007 #9
    OK then. Next question: Why did you decide do use it? How are you using it? Does it make sense?
  11. Mar 23, 2007 #10
    Okay, I got the answer.


    [tex]D^2=5625-1800t+144t^2+81t^2[/tex] *Note that it's +144t2, not -144t2 like before. Just a multiplication error.:uhh:






    Therefore, The ship will be the closest at 4 hours with the distance of 45km apart.
    Last edited: Mar 23, 2007
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