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Adding Vectors

  1. Aug 30, 2004 #1
    The velocity vector V1 has a magnitude of 3.0 m/s and is directed along the +x-axis. The velocity vector V2 has a magnitude of 2.0 m/s. The sum of the two is V3, so that V3 = V1+V2

    I was never shown how to add vectors and this question came up in class discussion and I was clueless.
     
  2. jcsd
  3. Aug 30, 2004 #2
    hmm what college is this you attend?

    jk

    nice name except one would excpet it in 1337

    form

    do u still need help with vectors?
     
  4. Aug 30, 2004 #3
    Yes well Eleet is the name of my clan on RS3:BA and our tag is 1337.

    and Yes I still need help!!
     
  5. Aug 30, 2004 #4

    chroot

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    If the direction of V2 is not given, then you can't add the vectors. You need both a magnitude and direction for each.

    If you have a magnitude and direction for each, just express each vector in terms of its components. For example, your vector V1 points along the positive x-axis and is 3 units long, so its components are:

    (3, 0)

    Read this as: "three units in the positive x-direction and zero units in the positive y-direction."

    If you have some other vector which points along the 45 degree line (measured increasing counter-clockwise from the positive x-axis) and is five units long, its components are

    [tex](5 \cdot \cos 45^o, 5 \cdot \sin 45^o)[/tex]

    as can be seen from simple trigonometry. To add two vectors with components (a, b) and (c, d), just add their components:

    (a, b) + (c, d) = (a+c, b+d).

    - Warren
     
    Last edited: Aug 30, 2004
  6. Aug 30, 2004 #5
    That is what I thought but the question does not give the direction of V2 which really confused me.

    Examples in the book gave directions w/the vectors.


    It also asked these True/False questions along with it:
    (T/F) The magnitude of V3 can be 2.0 m/s
    (T/F) The magnitude of V3 can be 0.0
    (T/F) The x-component of V3 can be 2.0 m/s
    (T/F) The magnitude of V3 can be 6.0 m/s
    (T/F) The magnitude of V3 can be 5.0 m/s
    (T/F) The magnitude of V3 can be -4.0 m/s
     
  7. Aug 30, 2004 #6

    chroot

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    Aha. You don't actually have to add the vectors to answer these true/false questions. Some of the choices are impossible, no matter what direction V2 points. Next time, please post the complete question at once so we'll be better able to help you.

    For example, the fourth choice "The magnitude of V3 can be 6.0 m/s," is impossible, because you only have a 3 unit vector and a 2 unit vector. If they point in the same direction, the best case, their magnitudes will add directly to 5 units. If they point in different directions, their sum will have a magnitude smaller than 5 units. If they point in opposite directions, their sum will only be one unit in length. There's no way to add a 3 unit and a 2 unit vector and get a 6 unit vector.

    The other wrong choices can be eliminated in the same way by the process of elimination. Let us know if you can't eliminate the others.

    - Warren
     
  8. Aug 30, 2004 #7
    Sorry about not posting the rest of the problem.
    But what you told me helped because I got them right after you explained it.

    Thankyou For teaching me something I did not know before.

    -Shaun
     
  9. Aug 30, 2004 #8
    "If you have some other vector which points along the 45 degree line (measured increasing counter-clockwise from the positive x-axis) and is five units long, its components are

    [tex](5 \cdot \cos 45^o, 5 \cdot \sin 45^o)[/tex]
    "

    This is often also just writen as

    [tex][\sqrt{\frac{25} {2}},\sqrt{\frac{25} {2}}][/tex]
     
  10. Aug 30, 2004 #9

    chroot

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    Of course, Extravagent, you can evaluate the sin and cos of 45 degrees easily. I chose to express it the way I did to make it clear how it was constructed.

    - Warren
     
  11. Aug 30, 2004 #10
    I have nothing against that. I was just commenting how it is usually writen in books. Your way is a good middle step for if the question asks to write a vector from the info: a vector of length 5 units, 45 degs above the + x-axis... While the way I showed it is often a way a vector is presented.
     
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