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Stationary object, constant velocity?

  1. Mar 2, 2014 #1
    I had an interesting debate with a friend of mine recently. They claim that a stationary object (we are JUST talking about our reference frame here) observed by us has a constant velocity, even if it is 0.

    However, I claim that since velocity is a vector quantity, which requires both magnitude and direction, we cannot attribute any velocity to a stationary object since it's direction is undefined. We can only state that its SPEED is zero (and also constant) as speed, being a scalar, is simply a numerical value which in this case does not change. I concede that one COULD argue that we simply take an XYZ graph (if working in 3D, for example), and simply state that each velocity component is zero (and constant). However, I claim that this is not a solution to the above argument but a crude workaround.

    A quick googling has unfortunately not revealed the answer, as I mainly get results from pages such as GCSE bitesize and the likes. :uhh:
     
  2. jcsd
  3. Mar 2, 2014 #2

    Drakkith

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    An object traveling with a constant velocity in the X direction still has velocity in the Z and Y directions. It's just zero. If it helps, consider an object moving half in the X direction and half in the Y direction. You can break the velocity vector into individual X and Y vectors, each with a different magnitude than the diagonal one.
     
  4. Mar 2, 2014 #3

    jtbell

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    The velocity of a stationary object, in Cartesian vector form, is ##\vec v = 0 \hat x + 0 \hat y + 0 \hat z##, where ##\hat x##, ##\hat y## and ##\hat z## are unit vectors in the x, y, and z directions.
     
  5. Mar 2, 2014 #4
    This issue is frequently brought up here, in somewhat different forms. The root cause is that the interpretation of a vector as some magnitude with direction does not work well for the zero vector. Still the zero vector is perfectly valid a vector, and is unique. Anything with zero speed in some frame has zero velocity in that frame.
     
  6. Mar 3, 2014 #5
    Aha, so it seems one of my suspicions was indeed correct. Thanks all for replying.
     
  7. Mar 3, 2014 #6

    A.T.

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    It's the other way around: Your argument is based on a crude definition of a vector, which doesn't account for the zero vector.
     
  8. Mar 3, 2014 #7
    All seems highly semantic to me.

    It's like saying, "Well there are no trees in the forest, but there are still trees because there are none of them."

    Much thought with little consequence.
     
  9. Mar 3, 2014 #8
    No, its more like saying that the number of trees in the forest is zero. Zero is a number just like any other.
     
  10. Mar 3, 2014 #9
    Yeah and what I mean is that it's about as much use as standing in the same empty forest and declaring there are zero dragons. In any situation you could write an infinite list of things that there are none of.

    As I said though, it's all just a bit of a semantic argument. When people refer to any value or amount as 'zero', we all know what we all mean. The philosophical discussion of 'actual existence' is kind of moot. Hence, as I also said, much thought with little consequence!
     
  11. Mar 3, 2014 #10

    sophiecentaur

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    Surely, the 'zero ness' attribute of zero is not inherently different from the 'one ness' of one or the 'two ness' of two. Arithmetical operations work with numbers alone and you don't need to declare 'trees' for the sums to be valid.

    We should leave sa1988's "much thought" to the Mathematicians, perhaps unless we know we're going to get anywhere with this.
     
  12. Mar 4, 2014 #11
    Tbh I seemed to have overcomplicated the situation. As far as I can tell, this is the simplest and most valid answer to my question.

     
  13. Mar 4, 2014 #12
    Yes, but we're talking about trees, not dragons. Someone asks "How many trees are there?", and someone else answers "there are zero trees". The statement "there are zero dragons" is true, but is not an answer to the question asked.
     
  14. Mar 4, 2014 #13
    Ok then.
     
  15. Mar 4, 2014 #14
    I agree with A.T. Your friend was right and you were wrong.

    Chet
     
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