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Acceleration, and displacement

  1. Mar 1, 2012 #1
    I think we can't have a negative displacement. Is that right?

    Acceleration: the change in displacement / change in time squared
    Since displacement can't be negative, and time can't be negative, acceleration also can't be negative.
    But I know (it's in the books) that when we have decreasing acceleration, it's negative. How is that?
     
  2. jcsd
  3. Mar 1, 2012 #2

    Doc Al

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    Staff: Mentor

    Why do you think this? Realize that whether a vector is negative or not just depends on the sign convention used when specifying its components.
    Better to think of acceleration as Δv/Δt.
    Again, the sign of a vector is rather arbitrary.
    Acceleration is a vector. A negative acceleration just means that the acceleration vector points in the negative direction. For example, if you take up as positive, the acceleration of a falling body will be negative.
     
  4. Mar 1, 2012 #3
    I understand now, it's the direction that is negative, not the magnitude.

    Why is it better to think of acceleration as Δv/Δt?
    I guess it just a matter of which is more intuitive/elegant, but maybe there is another reason; is there?
     
  5. Mar 2, 2012 #4
    Acceleration is the derivative of the velocity so it's dv/dt.

    The sign of the acceleration can have different criteria, I think an acceleration is negative if the scalar product (dv/dt)•v < 0 and positive if (dv/dt)•v > 0. If the scalar product it's zero then the acceleration is perpendicular to the motion. This is the case in circular motion, for example.
     
  6. Mar 2, 2012 #5

    Doc Al

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    Well, Δv/Δt is the definition of acceleration (at least average acceleration).

    Blindly using Δx/(Δt)2 can lead to silly results. Imagine something moving at a constant velocity of 10 m/s for 1 second. Δx = 10, Δt = 1. Obviously the acceleration is zero here, so that formula fails.
     
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