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Derivative of displacement with respect to time

  1. Mar 31, 2015 #1
    What's the difference between derivative of displacement with respect to time and derivative of position with respect to time ?I have read derivative of displacement with respect to time is velocity and derivative of position with respect to time is also velocity,so what's the difference?
    I think derivative of displacement with respect to time gives average velocity and derivative of position with respect to time gives instantaneous velocity.Is it right?
     
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  3. Mar 31, 2015 #2

    rcgldr

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    Update - the original post is confusing, derivative of position doesn't make sense. Displacement means a change in position, so the derivative of displacement is the same derivative of change in position. Speed is the derivative of total distance traveled versus time.

    Previously I thought the original post was trying to imply that displacement meant total distance traveled as opposed to change in position, so if displacement was redefined as total distance traveled, then by that redefinition, the derivative of change in total distance traveled versus time would be speed.
     
    Last edited: Apr 1, 2015
  4. Mar 31, 2015 #3
    Is it right?
     
  5. Mar 31, 2015 #4
    Are we talking about position vector (the vector that always points from the origin to the position of the particle )?
     
  6. Apr 1, 2015 #5

    jtbell

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    Yes, the position vector ##\vec r## is generally assumed to have its "tail" at the origin. A particle's instantaneous velocity vector is the derivative of the position vector: ##\vec v = d \vec r / dt##.

    Suppose that at some time ##t_0## the particle is located at position ##\vec r_0##. Then at some later time t, it is located at position ##\vec r## and its displacement from ##\vec r_0## is ##\vec r - \vec r_0##. This vector has its "tail" at ##\vec r_0##. The derivative of this displacement is $$\frac{d}{dt} (\vec r - \vec r_0)$$ What does this equal?
     
    Last edited: Apr 1, 2015
  7. Apr 1, 2015 #6
    But why?Displacement is a vector quantity and it is not the same as distance , it is the object's overall change in position.It is th shortest distance between initial and final point.If the body goes for 2 metre and then walk for 2 meter in revers direction it is again on the same location,so distance is 4 meter but displacement is zero.Considering this example displacement and position looks same.
     
  8. Apr 1, 2015 #7
    Let's consider the following case
    upload_2015-4-1_11-20-36.png
    displacement=change is position .I am not able to explore is there any case when
    displacement ≠ change is position
    Can you give me one example please?
     
  9. Apr 1, 2015 #8

    rcgldr

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    I updated my previous post. It's not clear to me if the issue is a definition of speed or an issue with the definition of displacement. In mathematics and science, displacement and a change in position are the same thing, so the original post is confusing. Speed is the derivative of total distance traveled versus time. Velocity is the derivative of displacement (which is the same as change in position) versus time.
     
    Last edited: Apr 1, 2015
  10. Apr 1, 2015 #9
    They are the same thing.
    The displacement is a difference between the current position vector [r(t), vector] and the initial position [r(0), vector].
    As the initial position is a constant, the derivative of displacement will be just the derivative of r(t).
     
  11. Apr 1, 2015 #10

    jtbell

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    I don't understand what this diagram is supposed to show. Can you please explain it?
     
  12. Apr 2, 2015 #11
    I'm a high school student, so correct me if I'm wrong, but this is my understanding:

    In terms of integrals, taking a velocity graph and integrating it to find the net area under a graph will yield displacement, since where velocity is negative (the opposite direction), it will count as negative area (movement in the opposite direction). This makes sense, when you think about it applied to life. However, finding the area between the velocity curve and the x axis will yield distance travelled, which is a matter of finding where the velocity is negative and making it positive on that interval when integrating. Distance and displacement are both derivatives of velocity, but the difference is that displacement is a vector, whereas distance travelled is a scalar quantity that doesn't rely on direction. If you imagine a sine wave representing velocity, on [0,2pi] displacement is zero whereas distance is 2. Think of distance in terms of integrating the absolute value of a function. Displacement will always be change in position, that is what it is defined as.

    I'm sorry if that's not what you were asking, I am a bit confused as to what the question asks.
     
  13. Apr 2, 2015 #12

    HallsofIvy

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    You are over complicating things by talking about integrating to begin with. If an object moves from, say, (0, 0) to (1, 2) its displacement is the vector from (0, 0) to (1, 1) which is normally represented by x+ 2j. If it did that is 3 seconds, it velocity is the vector (1/3)i+ (2/3)j. With that same movement, the distance it has moved is the distance from (0, 0) to (1, 2), [itex]\sqrt{5}[/itex] and its speed is [itex]\sqrt{5}/3[/itex]. "Displacement" and "velocity" are vectors. "Distance" and "speed" are numbers.
     
    Last edited: Apr 2, 2015
  14. Apr 2, 2015 #13
    Careful. You have it backwards in that part in red.
    Also, your integral example does not look right. For what expression of the sinusoidal velocity you get 2 for displacement?
    Distance traveled is not the same as "distance". The distance between two points is measured along the straight line connecting the two points. This last point is just a matter of being careful while using the terms.

    I think you understand the idea. Just have to be more careful when you try to explain it.
     
  15. Apr 2, 2015 #14
    Right. My bad. The magnitude of velocity is the antiderivative of position, while the vector of velocity is the antiderivative of displacement.

    I wasn't really thinking with what I said about the integral of sin(x), oops. Fortunately, the general idea was still conveyed, I believe (That the total area is greater than zero, while net area is zero)
     
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