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Problem with vectors

  1. Aug 4, 2007 #1
    1. The problem statement, all variables and given/known data
    We have two vectors: one with a speed of 3 m/s to the northwest, let's call this vector a, and we also have a vector b moving to the west with a speed of 5 m/s.

    Determine: a + b, a - b and a - 2b


    2. Relevant equations
    I have no idea.


    3. The attempt at a solution
    I tried to put the vectors in terms of unit vectors, but that didn't work, then I tried to use Pythagoras' Theorem, but that wasn't right either. I'm getting really frustrated with this problem! :cry:
     
  2. jcsd
  3. Aug 4, 2007 #2

    mjsd

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    you need a coordinate system so that you can put vectors into component form
    a = ( , ) ; b = ( , ) then addition becames simple. so first determine a set of axes and then give your vectors the appropriate coordinates before moving on. shall need some simple trig I think
     
  4. Aug 4, 2007 #3
    Could you show me how to do this?
     
  5. Aug 4, 2007 #4

    HallsofIvy

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    Basically, what you just said you tried. Since you don't show how you tried, I don't know why it "didn't work".

    "one with a speed of 3 m/s to the northwest, let's call this vector a, and we also have a vector b moving to the west with a speed of 5 m/s."
    Okay, so [itex]\vec{a}[/itex] has equal [itex]\vec{i}[/itex] and [itex]\vec{j}[/itex] components except that the [itex]\vec{i}[/itex] component is negative. Set it up as a right triangle with legs x and x, hypotenuse of length 3. Use the Pythagorean theorem to determine x. The vector is [itex]-x\vec{i}+ x\vec{j}[/itex].

    b is due west with "length" 5 so it should be easy to write it in [itex]x\vec{i}+ y\vec{j}[/itex] form!

    Once you have those two, the arithmetic is simple.
     
    Last edited: Aug 5, 2007
  6. Aug 4, 2007 #5

    mjsd

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    ok, let me give u an example.

    take the N direction as your +ve y-axis direction and E as your +ve x-axis direction. then a velocity vector pointing at S with magnitude 2m/s has a vector form based on this set of coord sys of
    v=(0,-2)
    and for a velocity vector pointing at say SW with magnitude [tex]\sqrt{2}[/tex] m/s has
    u=(-1,-1)
     
  7. Aug 5, 2007 #6
    I'm sorry I didn't show how I did it, but I don't have internet right now. I will show how I did it when I have internet.
     
  8. Aug 5, 2007 #7

    Kurdt

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    As everyone else has said you need to find the components of the vectors and then the calculations are fairly easy. In this situation south-north will be the y-axis and west to east will be the x-axis of a cartesian coordinate system. The unit vectors will then be [itex]\mathbf{\hat{i}} [/itex] and [itex]\mathbf{\hat{j}} [/itex]. You know the magnitude of the vectors so you can work out the components.

    [tex] |\mathbf{a}|=\sqrt{x^2+y^2} [/tex]

    For vector a you know that the x and y components must be the same and for vector b you know the y component is 0.

    Post your attempt when you get a chance.
     
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