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Unit Vector

  1. Oct 13, 2015 #1
    1. The problem statement, all variables and given/known dat

    A car travels 20 mi at 60 degrees north of west, then 35 mi at 45 degrees north of east.

    Express each displacement vector in unit vector notation. Take the +x-axis due east and the +y-axis due north. Use the component method to obtain the resultant displacement vector in unit vector notation.

    2. Relevant equations
    [tex]\ A = \ A_x i + \ A_y j[/tex]
    [tex]\ B = \ B_x i + \ B_y j[/tex]
    [tex]\ C = \ C_x i + \ C_y j[/tex]


    3. The attempt at a solution

    A= -20sin60i+35cos60j
    B=35cos45i+20sin45
    C=?

    How do I find C, since no lengths were given. Did I do A and B correctly?
     
    Last edited: Oct 13, 2015
  2. jcsd
  3. Oct 13, 2015 #2

    berkeman

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    Staff: Mentor

    The solutions look incorrect so far. For A, both the x and y components involve the 20 mile displacement, and for B, both the x and y components involve the 35 mile displacement. Also, can you show a sketch that shows how you determine when to use the cos() and when to use the sin() to find the components on the x & y axes?

    Finally, you are indeed given the lengths of 20 and 35 miles, so you will be able to calculate the sum vector C = A + B. :smile:
     
  4. Oct 13, 2015 #3

    berkeman

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    Also, you can't reduce the vector components down to a single scalar number "7.5". Vectors have a magnitude and a displacement, and can be expressed as that two number combination, or in rectangular component form. Are you familiar with how to convert between those two forms?
     
  5. Oct 13, 2015 #4
    Oh yes you are right. But are the sin and cos right?

    I do not know to convert to rectangular component form. I think we are suppose to put it in unit vector notation.
     

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  6. Oct 13, 2015 #5

    berkeman

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  7. Oct 18, 2015 #6
    Okay I redid everything and found:

    A=20cos60i+20sin60j
    B=35cos45i+35sin45j
    C=(-20cos60+35cos45)i+(20sin60+35sin45)j

    Did I do all the trig stuff right? I am suppose to find the magnitude too which I found was 97.62.
     
  8. Oct 19, 2015 #7
    Bump.
     
  9. Oct 20, 2015 #8

    berkeman

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    I believe there is a sign error in this equation... :smile:
     
  10. Oct 20, 2015 #9

    Mister T

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    Would it make any difference in your solution if the direction of ##\mathrm{\vec{A}}## were instead 60° north of east?
     
    Last edited: Oct 20, 2015
  11. Oct 20, 2015 #10
    Oh I fixed it. I think I understand everything now. But I want to make sure, C is the resultant right? Because C is technically R.

    Yes because that would change the problem entirely.
     
  12. Oct 20, 2015 #11

    berkeman

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    Could you show your work for the final answer for C? Thanks, that will help us answer...
     
  13. Oct 20, 2015 #12

    berkeman

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    He's just seconding my comment about the sign error... :smile:
     
  14. Oct 20, 2015 #13

    Mister T

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    The problem statement doesn't specify. It was you who chose ##\mathrm{\vec{A}}## to represent one of the two given vectors and ##\mathrm{\vec{B}}## the other. Likewise you are free to choose ##\mathrm{\vec{C}}## or ##\mathrm{\vec{R}}## as their sum.

    If ##\mathrm{\vec{C}}## is your choice, then you'd write ##\mathrm{\vec{A}+\vec{B}=\vec C}##.

    If ##\mathrm{\vec{R}}## is your choice, then you'd write ##\mathrm{\vec{A}+\vec{B}=\vec R}##.

    The term "resultant" has the same meaning as "vector sum".
     
  15. Oct 20, 2015 #14

    Mister T

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    No, you are not supposed to find the magnitude. You are supposed to express the result in unit vector notation.
     
  16. Oct 20, 2015 #15
    There was no C. I figured that out. I was confused because in his example he kept saying R=A+B+C. But since I was only given two vectors it would be R=A+B. I found R=14.75i+42.07j. The magnitude is 44.58 mi at [itex]70.68^o[/itex].
     
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