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Having a little trouble, very typical question

  1. Aug 26, 2006 #1

    DB

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    im not use to working is 3D so i need some help

    you are going to a restaurant. you leave your apartement and take the elevator 30 m down, then 15 m south to the exit. then u travel 200 m east and then 100 m to the restaurant. what is your displacement.

    so what i did is:

    -30k
    -15j
    +200i
    +100j

    then after adding them up:

    [tex]\Delta d=\sqrt{200i^2+85j^2+(-30k^2)}[/tex]

    displacement = ~ 219.37 m

    is this right? and how do i represent the final direction?

    thanks
     
  2. jcsd
  3. Aug 26, 2006 #2

    quasar987

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    You might want to double check Pythagora's thm there! It says


    [tex]\Delta d=\sqrt{\Delta x^2+\Delta y^2+\Delta z^2}[/tex]
     
  4. Aug 26, 2006 #3

    DB

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    sry, i cant find the mistake your talking about, would that 85 be -115?
     
  5. Aug 26, 2006 #4
    That's an interesting one DB. I just checked the arithmetic and you do have the right answer in terms of distance. But whilst I can know how to use tangents to work out the horizontal direction in degrees that you ought to supply in addition to the distance, I'm not familiar with using this to express the downward component. Perhaps one of the more learned chaps can chip in here.
     
  6. Aug 26, 2006 #5

    quasar987

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    Firstly, how do we know in which direction the last 100m is?
     
  7. Aug 26, 2006 #6

    quasar987

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    I was refering to the fact that you did not square the distances. I.e. you used the equation

    [tex]\Delta d = \sqrt{\Delta x+\Delta y+\Delta z}[/tex]

    Edit: Oh i understand! By 200i², you mean (200i)². The parenthesis are important because [itex]200i^2[/itex] means (200)*(i²)=200.
     
    Last edited: Aug 26, 2006
  8. Aug 27, 2006 #7

    DB

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    oh im sorry, it is suposed to say 100m North
     
  9. Aug 27, 2006 #8

    HallsofIvy

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    I found myself not certain as to whether "displacement" was a vector (my first choice) or just the distance (which several of the answers are assuming) so I went to "google" to check. No luck! What I got was "A vector or the magnitude of a vector from the initial position to the final position"! In other words, it could be either and I think you had better check exactly which meaning it is given in your class.

    In any case: taking the "i" vector to the east, the "j" vector north, the "k" vector upward, you: "take the elevator 30 m down": -30k; "then 15 m south to the exit": -15j;"then u travel 200 m east": 200i; "then 100 m (NORTH)to the restaurant" (North added from your later respons): 100j.
    Adding all those, your displacement vector is -30k- 15j+ 200i+ 100j= 200i+ 85j- 30k. Your numerical displacement (distance) is
    [tex]\sqrt{(200)^2+ (85)^2+ (-30)^2}= \sqrt{40000+ 7225+ 900}= \sqrt{48125} m[/tex].
     
    Last edited: Aug 27, 2006
  10. Aug 27, 2006 #9
    You got the distance portion right. To get the direction you need to know use tangents. You know that you moved 85m North and 200m East. You can express these as an opposite over adjacent ratio, and find the bearing in degrees using the INV tangent function on this here online calculator.

    http://www.math.com/students/calculators/source/scientific.htm

    You can then repeat with 30/sqrt(852 + 2002) to get a declination.

    But I suspect that the answer they were looking for was just the 219.37m distance.
     
    Last edited: Aug 27, 2006
  11. Aug 27, 2006 #10

    DB

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    thanks everyone, btw hallsofivy displacement is a vector, distance is not. just to cleat that up :) thanks again i understand the problem now
     
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