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Why is acceleartion in denominator ^2 units?

  1. Mar 30, 2005 #1
    Let's say i have some function S(t) where t is in seconds and S(t) gives feet per second.

    As I take the derivatives of S to get the acceleration function, why is the feet unit squared? What's the best way to conceptualize what's going on?
  2. jcsd
  3. Mar 30, 2005 #2
    you have it a little mixed up.
    S(t) would give you position not velocity
    it's derivative S'(t) would give you ft/sec
    an the derivative of S''(t) would give you acc. which is ft/(sec)^2

    The first derivative tells you how fast your position changes with time, the second derivative is telling how fast your first rate of change is changing with respect to time, hence ft per seconds squared.
  4. Mar 30, 2005 #3
    It's the definition of acceleration, ie. rate of change of velocity. Velocity is rate of change of position (m/s), so rate of change of velocity is m/s/s.
  5. Mar 30, 2005 #4


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    So no feet is getting squared.On the other hand,the second does...

  6. Mar 30, 2005 #5
    That's the best explanation i've seen. Thanks.

    And yeah that should be seconds squared. Also, my plural use of deriviative was in respect to s''.
  7. Mar 30, 2005 #6
    Distance is in meters

    Velocity is in meters per second

    Acceleration is in meters per second per second.

    In each step you are dividing by seconds.

    Velocity / Seconds = [tex] \frac{\frac{m}{s}}{s}} = \frac{m}{s} * \frac{1}{s} = \frac{m}{s^2} [/tex]

    Likewise jerk is the change in acceleration, dividing by seconds:

    Acceleration / Seconds = [tex] \frac{\frac{m}{s^2}}{s}} = \frac{m}{s^2} * \frac{1}{s} = \frac{m}{s^3} [/tex]
  8. Mar 30, 2005 #7
    i never knew there was something after acceleration :|
  9. Mar 30, 2005 #8
    Theres infinitely many, they only named the first five I think.
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