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Finding closest distance

  1. Jan 16, 2007 #1
    1. The problem statement, all variables and given/known data
    Ship A, sailing due east at 8 km/h, sights ship B 5km to the southeast when ship B is sailing due north at 6km/h. How close to each other will the two ships get?


    2. Relevant equations



    3. The attempt at a solution
    I drew the picture. Ship a distance is 8t and B is 6t

    Use the a^2+b^2=c^2
     
  2. jcsd
  3. Jan 16, 2007 #2

    Dick

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    Seems to me there are two coordinates of interest here. Hint.
     
  4. Jan 16, 2007 #3
    sorry i'm lost
     
  5. Jan 16, 2007 #4

    HallsofIvy

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    "Ship a distance is 8t and B is 6t" distance from what? Set up a coordinate system, say with A initially at the origin. What are the coordinates of ship A at time t? What are the initial coordinates of ship B? What are the coordinates of ship A at time t? What is the (square of the) distance between those points as a function of t?
     
  6. Jan 17, 2007 #5
    ok [tex]D^2=(-3.53+6t^2)+8t^2 dD/dt=(200t-42.36)/whatever\\0=200t-42.36\\t=0.212\\[/tex]
    I plugged that in the distance formula and got a wrong answer
     
    Last edited: Jan 17, 2007
  7. Jan 17, 2007 #6

    Dick

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    Follow HallsOfIvy's suggest and write down the position of the two ships in xy coordinates as a function of time first. It's difficult to tell where you went wrong from what you post.
     
  8. Jan 17, 2007 #7
    yes I have a drawing. I used the sin law to find the distance of the 2 sides.

    Coordinates (3.53,0) and (3.53,-3.53)

    By the way how do you put spaces in latex coding I tried \\ it didnt work
     
  9. Jan 17, 2007 #8

    arildno

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    Eeh??

    Do you even know what coordinates are??

    Considered as a function of time, what is ship A's position measured from an origin lying where A was, and sighted B somewhere at t=0?
    And, with the same choice of origin, what is B's position as a function of time?
     
    Last edited: Jan 17, 2007
  10. Jan 17, 2007 #9

    Dick

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    At what time? Where is the time dependence? (Not sure about your latex question, sorry).
     
  11. Jan 17, 2007 #10
    yes those are the coordinates.from the origin (0,0) using pythagorean theorem,

    i use 5sin45=a=3.53, b=3.53
     
  12. Jan 17, 2007 #11

    arildno

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    1. What choice have you made of positive axes?
    2. What was A's position at t=0?
    3. What is A's position as a function of time?
    4. What is B's position as a function of time?
     
  13. Jan 17, 2007 #12
    A's position at t=0 is (0,0)
    A's position as a function of time is (3.53-8t,0)
    B's position as a function of time is (3.53,-3.53-6t)
     
  14. Jan 17, 2007 #13

    arildno

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    Is (3.53-8*0,0)=(0,0)?? :confused:?
     
  15. Jan 17, 2007 #14
    ah crap....
    A is (8t,0)
     
  16. Jan 17, 2007 #15
    **** or is it
    A(8t,0)
    B(3.53,-3.53+6t)

    My guess is that's right if not I quit school
     
  17. Jan 17, 2007 #16

    Dick

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    You can stay in school! Now what is D^2 as a function of t?
     
  18. Jan 17, 2007 #17
    D^2=(-3.53+6t-0)^2+(3.53-8t)^2

    I did derivative. Set it to 0. Got 0.7

    Thats not the answer

    I'm suppost to plug time into the D^2 and get the distance correct?
     
  19. Jan 17, 2007 #18

    Dick

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    You seem to be trying to do the right thing. What is the derivative of D^2?
     
  20. Jan 17, 2007 #19
    I got it to be:

    After the expansion and derivative

    dD/dt=200t-98.84/(the rest here doesn't matter cause i'm setting other side to zero)
     
  21. Jan 17, 2007 #20

    Dick

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    All seems ok. What is the answer supposed to be?
     
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