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Homework Help: Adding & Subtracting Vecots And Relative Velocity

  1. Apr 12, 2005 #1
    Adding & Subtracting Vectors And Relative Velocity

    Hey everyone, I am so confused with this it's not funny. This is my first time accessing a site like thisss, but I am desperate seeing as though it's school holidays atm where I am and I can't talk to my physics teacher for help. Anyhow, I hope one of you guys can help me out. :smile:

    Ok I have 2 problems.

    One is with adding and subtracting vectors. Ok when I have to vectors to add or subtract say

    12ms-1 west + 14 ms-1 north

    <IMG SRC="http://members.optushome.com.au/rossverg2002/1.JPG"> [Broken]

    Before I do anything I draw my diagram as descbried above. The problem is that in oder to add them I need to have the vectors touching arrow-to-tail. The problem is which one of the arrows do I reverse and how do I express the new velocity (with a - sign?). Does it even matter which arrow (velocity and direction) I reverse?

    Ok second problem :frown: ... I am confused with relative velocity. In my textbook there is a formula:

    Velocity of a relative to b = velocity of a - velocity of b.

    How do I know when to use the formula or vector diagrams? Is it only meant for velocities in a straight line? What do I do when there are 2 dimensional velcoities (like a velocity north relative to a velocity east); do I use vector diagrams instead of teh formula??

    Here is an example I am stuck on with the above problem.

    Car A travels at a velocity of 30.5 ms-1 north while car B travels at a velocity 25 ms-1 east. What is the velocity of B relative to A?

    If I use the formula, I get 55.5 ms-1. (Velocity of b rel to a = 25 - (-30.5) which = 55ms-1)

    If I use a vector diagram and find the velocity via pythagoras, I get 39.44 ms-1 N34degreesW.

    Which way is the right way?? :S:S:S

    Thankyou to all who help.
    Your help is greatly appreciated. o:)

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  2. jcsd
  3. Apr 12, 2005 #2


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    First, vectors have two important characteristics. They have a magnitude and they have a direction. The location isn't that important. If two vectors start from different origins, but have the same direction and magnitude, then they are equal.

    Going back to basic algebra:

    If A=B, and A+C=D, then what does B+C=? Obviously it equals D.

    Since the origins of your two vectors occupy the same spot, they're not set up to add together very well. Instead, you simply substitute in a new vector at the location you need that has the same magnitude and direction as one of your vectors. (In other words, you don't reverse the direction of either vector).

    Addition is commutative, so you don't care which vector you move (or, technically, substitute). You can add the vectors in either order.

    Try it both ways. You'll find you've created a parallelogram (a rectangular parellelogram for the example you gave). The sum equals the diagonal of the parallelogram either path you take. (That's why it's sometimes called the parallelogram law of addition).
    Last edited by a moderator: May 2, 2017
  4. Apr 12, 2005 #3


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    The second is the correct answer, almost. I think you made a typo on the direction, since it's 39.34 degrees West of North (N39W).

    You misinterpreted what they meant by the formula. If North is your positive x-axis (prinicipal direction) then east is your positive y-axis.

    Vector A = [30.5 , 0]. Vector B = [0 , 25].

    You subtract by components: Subtract the x-components from each other, subtract the y-components from each other to get a new vector.

    In other words, the new vector is [30.5-0 , 0-25], which equals [30.5, -25]

    You use the Pythagorean Theorem to find the magnitude and the tangent to find the direction.
  5. Apr 12, 2005 #4

    Doc Al

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    You may find this discussion of vector addition helpful: http://www.glenbrook.k12.il.us/gbssci/phys/Class/vectors/u3l1b.html [Broken]
    Last edited by a moderator: May 2, 2017
  6. Apr 12, 2005 #5
    Thanks Bob G so much for clearlying that up.
    I've got a much clearer picture of what this is all about now.

    Thanks DocAl for that link as well. :)
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