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Vector 2D motion Question.

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Sam Walks: 100m (S30W) + 200m (N30E) + 50m (N) Find sam's displacement using a diagram aswell as mathematically.


    2. Relevant equations
    b(sq) = a(sq) + c(sq) - 2acCOSB
    a(sq) + b(sq) = c(sq)

    3. The attempt at a solution
    Okay I drew my graph, Scaled it 1cm = 50m
    So I first drew 100m (S30W) (2cm) and than I go to draw 200 (N30E) but It looks like i have to overlap 100m S30W? Please Explain to me how I am suppose to draw this thing out. do I overlap or do I draw beside it? I don't know.
     
    Last edited: Sep 29, 2012
  2. jcsd
  3. Sep 29, 2012 #2
    Can you show us the diagram.
     
  4. Sep 29, 2012 #3
    It's just a on a cartesian plane marked North South East and West. The only line I drew was in the SW quadrant. I only drew a 2cm line with an angle of 30 degrees from south. Line started at the origin.
     
  5. Sep 29, 2012 #4
    Now what about the second leg of the journey?
    Remember the second leg starts from end of first leg which means you have new origin.
     
  6. Sep 29, 2012 #5
    Ohh okay so I'll measure 30 degrees NE from the end of the first line and then north for the third step from the 2nd line.

    EDIT: That makes no sense though, Still with the new origin, I'll be overlapping the the last line.
     
  7. Sep 29, 2012 #6
    That's correct.
    So where is Sam in relation to the FIRST origin?

    Yes he is backtracking. Sam cannot be in 2 places at one time.
     
    Last edited: Sep 30, 2012
  8. Sep 30, 2012 #7
    Sam should be 100m into the NE Quadrant. Now I have to go North. That gives me two side lengths. Thanks, I get it now. But Wait, I have another small question. What does it mean when a question says resultant velocity? Is it different from displacement?
     
  9. Sep 30, 2012 #8
    I'm sure what really you mean.
    Surely displacement is different from resultant velocity.

    Velocity as well as displacement are vectors. Result or we call it resultant is the net vector after vector operations(addition, multplication and subtraction)
     
    Last edited: Sep 30, 2012
  10. Sep 30, 2012 #9

    ehild

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    You walk 100 m forward in some direction and then you walk 200 m backward how far you are from your starting point?

    A figure would be nice to make yourself understand. See attached one, what is the distance of point C from the origin? What is the angle OC encloses with North?

    ehild
     

    Attached Files:

  11. Sep 30, 2012 #10
    I figured the unknown side out and it came to about 7.1cm. I will have to do it mathematically to make sure I'm correct. Now, Do you mean the space inbetween North and the 7.1cm line? I think you may have to use the Sine Law for that. but I'm not quite sure how to actually use it to get that angle. I know there is a Z pattern but I don't know if that applies here.
     
  12. Sep 30, 2012 #11

    ehild

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    Try to find the length of OC mathematically. You can apply either the Law of Cosines, and for the angle, you can use Sine Law, or find the x,y components of the vectors and add them by components. Then find both the length and the angle of vector OC using its components.


    ehild
     
  13. Sep 30, 2012 #12
    How am I suppose to find side OC when I only have two side lengths and not a known angle ?
    To do the cosine law you need an angle so you can do "COS(angle here)" You need an angle for the Sine Law as well and since I don't have one, I can't do it?

    What do you mean by x and y components of the vectors?
     
  14. Sep 30, 2012 #13

    ehild

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    You know the angle of OB with respect to the North direction (30°), and from that you know the angle between OB and BC, as BC points to North.

    Haven't you learned about components of vectors?

    ehild
     
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