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Change in Velocity

  1. May 29, 2012 #1
    1. The problem statement, all variables and given/known data
    A car's velocity changes from 13.9 m/s east to 12.8 m/s northeast. What is the change in it's velocity?
    It's a 4 marker.

    2. Relevant equations



    3. The attempt at a solution
    Honestly, I don't really know how to start this. I know it's a vector difference and I've drawn the diagram which isn't too tough. The fact that I'm using a negative vector is throwing me off.

    Could anyone nudge me in the right direction?
     
  2. jcsd
  3. May 29, 2012 #2
    How are you using a negative vector?
     
  4. May 30, 2012 #3
    Vector subtraction?

    v2-v1= v2+(-v1)

    So does the 13.9 m/s here, not turn into a negative value?
     
  5. May 30, 2012 #4
    It's the length and direction of the vector that goes from the first velocity to the second velocity. The difference vector isn't so much 'negative' as merely heading in a different direction.
     
  6. May 30, 2012 #5
    If you have studied about vectors and unit vectors then try finding unit vectors along north and north east.
    Unit vectors have zero magnitude of 1 and a direction
    So, vector(v) = magnitude(v) * direction (v)
    Or, vector(v) = magnitude(v) * unit vector (v)

    Just subtract the two vectors you get to get the final vector, and then find its magnitude.

    PS: you wil always get a positive change as magnitude of a vector is always positive.
     
  7. May 30, 2012 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You subtract. Subtract vectors. One way to do that is to write each vector in north and east "components"- ie < 13.9, 0> for the first vector and <a, b> for the second vector where [itex]\sqrt{a^2+ b^2}= 12.8}[itex] and, because "northeast" has equal parts "north" and "east", a= b.

    Alternatively, you could do this "trigonometrically". Represent each velocity vector as a line, "13.9 m/s east" of length 13.9 to the right, and "12.8 m/s northeast" of length 12.8 making an angle of 45 degrees with the first line. The "difference" is the third side of that triangle.
     
  8. May 30, 2012 #7
    Ahh so "Northeast" represents an exact angle of 45 degrees? I had a hunch of this, but couldn't find it anywhere. That helps a lot!

    Thanks everyone else for your help too!
     
  9. May 31, 2012 #8
    yes NE means 45 angle with both north and east direction,
     
  10. Jun 17, 2013 #9
    Something about the question is throwing me off...

    Using vector subtraction I get 62° (61.8°) North of East -(Xtotal =4.85 m/s East & Ytotal =9.05 m/s North)
    Using trigonometry I get 62° (61.8°) North of West

    I think the second answer is right (North of West), not sure though.
    Can anyone help explain what I'm doing wrong?
     
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