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Vector Displacement and Magnitude

  1. Oct 14, 2009 #1
    A dog walks 3.50m south, then 8.20m at an angle 39.4 degrees north of east, and finally 15.0m west.

    a) What is the magnitude of the dog's total displacement

    b) What is the directionof the dog's total displacement where directly east is taken as zero degrees and counter-clockwise is positive?


    I have found components of some of the vectors

    Ax= 0 Ay= -3.50
    Bx= 15? (I am not sure about this) By= -3.50
    Cy= 0 Is Cx = 15?


    Then I used the formula C= sqrt Cx^2 + Cy^2
    Resulting in 15 meters as the magnitude of displacement. Which doesn't sound right to me since he must have passed his starting point along his route of 15 meters west.

    I then Used the formula tan -1 (Cy/Cx) resulting in 0 for his direction.

    Can someone please help?!?
     
  2. jcsd
  3. Oct 14, 2009 #2

    rl.bhat

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    You have not taken not account the second displacement. Resolve that into two components. That gives you the position of the second destination from x-axis. From that point reach the final destination and note down the position of the final destination.
     
  4. Oct 15, 2009 #3
    Would I use the formula C=sqrt(Ax+Bx)^2+(Ay+By)^2

    Resulting in a displacement magnitude of 16.55
     
  5. Oct 15, 2009 #4

    rl.bhat

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    Bx = 8.2*cos39.4.
    By = ...?
    Final position from y-axis is 15 - Bx
    Final position from x axis is ....?
     
  6. Oct 15, 2009 #5
    Were my previous components correct? Or do I find them by taking Ax= Acos(angle)
    Ay=Asin(angle)

    The angle for A would be -90 since he is walking directly south resulting in components of
    Ax=0
    Ay=-3.50

    Vector B is 39.4
    Bx=6.34
    By=5.20

    Don't Ay and By have to be the same since the tail of vector B starts and the tip of vector A?

    My components for C would then be
    Cx=6.34
    Cy=1.7

    Is this correct?
     
  7. Oct 15, 2009 #6

    rl.bhat

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    No.
    Cx = -15 + 6.34
    Cy = 1.7
     
  8. Oct 15, 2009 #7
    The final position from the y-axis would be 8.66
    The final position from the x-axis 3.50+1.7=5.2?

    So in vector notation the displacement is C=-8.66i + 1.7j

    To find the magnitude of the dogs total displacement I would use the equation

    C=sqrt(Ax+Bx)^2+(Ay+By)^2 Which equals 6.56

    And the direction of the dogs total displacement, where directly east is taken as zero degrees and counter-clockwise is positive, I would use the equation

    tan (Cy/Cx) Which equals -.003??? Do I use tan-1(Cy/Cx) Which equals -11
     
    Last edited: Oct 15, 2009
  9. Oct 15, 2009 #8
    Ok, so

    Cx=-8.66
    Cy=1.7 (I knew that I don't know why I put 5.2)

    For the direction I then have -.003 and -11 degrees for answers I am assuming I should use tan-1. How do I know when to use tan-1 or tan?
     
  10. Oct 15, 2009 #9
    Final answers......

    a) 6.56m

    b) -11 degrees

    ?????????????????????????????
     
  11. Oct 15, 2009 #10

    rl.bhat

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    How did you get 6.56?
    Angle should be in the counterclockwise. So....?
     
  12. Oct 15, 2009 #11
    To find the magnitude I have the equation C=sqrt(Ax+Bx)^2+(Ay+By)^2

    Which equals 6.56
     
  13. Oct 15, 2009 #12

    rl.bhat

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    From the starting point Cx = 8.66 and Cy = 1,7
    Find the resultant of the these two components.
     
  14. Oct 15, 2009 #13
    I am not sure what to do.

    Do I add the two components? 10.36?

    Or

    C=sqrt(Cx+Cy) Isn't Cx=-8.66? So you cant take the sqrt.

    I am confused
     
  15. Oct 15, 2009 #14

    rl.bhat

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    Cx and Cy are perpendicular to each other.So

    C = sqrt(Cx^2 + Cy^2)
     
  16. Oct 15, 2009 #15
    So, final answers....

    a)8.83m

    b)11 degrees
     
  17. Oct 15, 2009 #16

    rl.bhat

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    No. The angle is in the second quadrant.
    θ should be ......?
     
  18. Oct 15, 2009 #17
    168 degrees?

    Is the magnitude correct; 8.83?
     
  19. Oct 15, 2009 #18

    rl.bhat

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