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What does 'velocity is derivative of distance' mean?

  1. Jul 12, 2015 #1
    What does v equals dx/dt mean? I interpret v as: the limiting value as a vanishingly small value for time t (dt) goes to 0. Or lim as dt-->0 of dx/dt.
     
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  3. Jul 12, 2015 #2
  4. Jul 12, 2015 #3

    Drakkith

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    I believe that would be instantaneous velocity. In general, if the equation for position vs time is X= T2 +B, where B is a constant and T is the variable for time, then the equation for the velocity is the derivative: V = 2T
     
  5. Jul 12, 2015 #4
    Thank you for the reply
     
  6. Jul 13, 2015 #5

    Philip Wood

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    Velocity is a vector, defined as the derivative with respect to time of another vector: displacement, r, (from a given point). The idea is that we take a time interval, [itex]\Delta t[/itex], centred on the particular time instant, t, that we're interested in, and consider [itex]\Delta \mathbf r[/itex], the change in r over the time interval [itex]\Delta t[/itex]. The mean velocity over [itex]\Delta t[/itex] is then defined by
    [tex] \mathbf{v_{mean}} = \frac{\Delta \mathbf{v}}{\Delta t}.[/tex]
    The velocity, v, at time t is defined as the limit of the mean velocity over [itex]\Delta t[/itex], as [itex]\Delta t[/itex] tends to zero.
     
  7. Jul 13, 2015 #6

    CWatters

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    In plain English.. The slope of the graph of distance vs time.
     
  8. Jul 13, 2015 #7

    Philip Wood

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    Fine if v is the component of velocity in a given direction, but if v is velocity in its general, vector, sense, it has components in three dimensions, so it's hard to see what single graph it would be the slope of.
     
  9. Jul 13, 2015 #8

    Drakkith

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    Hmmm. Perhaps it would be the derivative of the magnitude of the displacement wrt time, which you would then multiply by the displacement's unit vector?
     
  10. Jul 14, 2015 #9
    Absolutely not - this would imply that velocity is always directed exactly away from the origin.
     
  11. Jul 14, 2015 #10

    robphy

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    If you have components of ##\vec v##, then those components are slopes in the corresponding position-coordinate-vs-time graphs. In three-dimensions, you have three such graphs.
    If you want to visualize ##\vec v## vectorially in a single "graph", you can visualize it in space (or spacetime) as certain tangent-vectors to the trajectory (or worldline).
     
  12. Jul 15, 2015 #11

    Philip Wood

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    I assumed Drakkith intended to say: plot distance along path against time, so slope of graph gives speed, then multiply by instantaneous unit vector in the direction the body is moving (that is by the magnitude of the velocity vector).
     
  13. Jul 15, 2015 #12

    vanhees71

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    It is indeed very important to first make yourself clear that velocity is a vector (its magnitude is a scalar and called speed; English is great in being precise enough to distinguish these concepts clearly). The use of vectors is of great help in physics from day 1 on anyway. So it's good to learn about it.

    One first has to start with a space-time model. In the introductory physics course you start with the (apparently) most simple successful space-time model, the Newton-Galilei space-time. According to this model, space and time are independent of any physical processes, and any observer describes space as a three-dimensional Euclidean space of points and time as an orientied one-dimensional continuum. You can visualize this space-time as a straight line (depicting the real numbers in the usual way) along which in each point there is a three-dimensional Euclidean space.

    You can discribe the three-dimensional Euclidean space by arbitrarily singling out one point ("the origin"). Then each point can be addressed by a vector, i.e., an arrow pointing from the origin to that point, the position vector ##\vec{x}##. The origin itself is given by the null vector. Using three rigid rods of unit length, perpendicular to each other, giving three vectors ##\vec{e}_j##, you have a Cartesian basis, and you can describe each point by the three components of the position ##\vec{x}=x^j \vec{e}_j##, where ##x^j## are real numbers, and ##j## runs from 1 to 3. If in an equation 2 indices (one upper, one lower) are the same, you understand that one takes the sum of the corresponding three expressions:
    $$\vec{x}=x^j \vec{e}_j \equiv \sum_{j=1}^3 x^j \vec{e}_j.$$
    Now the most simple system is a "point particle", which is a small macroscpic body (small compared to the typical scales of its motion; e.g., in astronomy at first approximation you can consider the Sun and the Earth as point particles orbiting each other due to gravitational interaction). It's position is given as a function of time ##\vec{x}=\vec{x}(t)##.

    Then the average velocity between two times ##t_1=t## and ##t_2=t+\Delta t## is defined as
    $$\langle \vec{v} \rangle_{\Delta t}=\frac{\vec{x}(t+\Delta t)-\vec{x}(t)}{\Delta t}.$$
    The meaning is immideately clear: The difference of two vectors in the numerator gives the arrow (displacement) of the body from its initial position ##\vec{x}(t)## to its final position ##\vec{x}(t+\Delta t)##, and dividing by the time ##\Delta t## it took to make this change of position, gives you a precise measure of how fast the particle is going and, also pretty important, from where to where it precisely went.

    Now you can let ##\Delta t \rightarrow 0##. Then you get the momentaneous velocity, which tells you the direction and magnitude of the change of position during an infinitesimal time interval ##\mathrm{d} t##. This is called a derivative with respect to time, i.e.,
    $$\vec{v}(t)=\lim_{\Delta t \rightarrow 0} \frac{\vec{x}(t+\Delta t)-\vec{x}(t)}{\Delta t}=\frac{\mathrm{d}}{\mathrm{d} t} \vec{x}(t)=\dot{\vec{x}}(t).$$
     
  14. Jul 15, 2015 #13

    Fredrik

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    dx/dt is a notation for x'(t), which by definition is equal to ##\lim_{\Delta t\to 0}\frac{\Delta x}{\Delta t}##. So you shouldn't put a "lim" in front of dx/dt.

    This is just a comment about the notation dx/dt. Everything else is explained by vanhees71 in the post above this one.
     
  15. Jul 15, 2015 #14
    Ah yes, I think I misinterpreted what he was saying - sorry Drakkith!
     
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