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    sunny0001's Avatar
    sunny0001 Posts: 1, Reputation: 1
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    #1

    Jan 26, 2010, 03:40 AM
    How displacement is a vector quantity
    How displacement is a vector quantity
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Jan 26, 2010, 08:59 AM

    Vectors have both magnitiude and direction. It's obvious that for displacement the magnitude is important (2 miles is definitely different than 3 miles), but so is direction - if you travel 3 miles to the east you end up in a different place that if you travel 3 miles to the west. So both magnitude and direction are important when talking about displacement. Hence displacement is a vector.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #3

    Jan 26, 2010, 09:02 AM

    What is the definition of displacement?

    It's the length of the straight path, from the starting point to the finish point of a body that moved (or is moving). The direction of that path is included.

    EDIT: Didn't see your post ebaines :o
    leechaolan's Avatar
    leechaolan Posts: 11, Reputation: 2
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    #4

    Feb 9, 2010, 07:40 PM

    Displacement is an entirely empirical concept and vectors (position vectors to be exact) were just used solely to model displacements. Vectors are really axiomatic based on our empirical evidence of how displacements work. After some time, these same vectors were used to model forces (which work in the same mathematical way), velocity accelerations and other time derivatives of displacements.

    Moreover, the concept of a vector itself is actually more abstract and more purely mathematical than that of a displacement. (if you study linear algebra, you'll see why) But as far as what displacements are, Unknown008 pretty much cleared things up on it's empirical nature. Displacements are the basis for the foundations of the axioms for vector arithmetic. Further on, two new axioms are used to define another mathematical object called a vector space which is significant in all modern physics. (that's past Lagrange's reformulation)

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