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Instantaneous velocity

  1. Sep 20, 2004 #1
    This is an equation of a position v time graph:

    x=342t^4-127t^3+1.87t^2+2.45t

    I need to use this expression to find the instantaneous velocity at t=0.000s
     
  2. jcsd
  3. Sep 20, 2004 #2

    arildno

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    And how is inst. velocity related to position?
     
  4. Sep 20, 2004 #3
    take the derivative of the first equation
    then just plug in t
    and volla you got instantious velocity
     
    Last edited: Sep 20, 2004
  5. Sep 20, 2004 #4
    x`=1368t^3-381t^2+3.74t+2.45

    is that correct?
     
  6. Sep 20, 2004 #5
    When you are done check your answer

    This is what i got
    its in white so you must highlight
    [tex] 1368x^3 - 381x^2+3.74X+2.45 [/tex]
    =2.45=instant velocity

    lol
    oh t is 0
    well this is how you would find instant acceleration at other times
    [tex] 4104x^2-762x+3.74[/tex]

    sorry I added extra but I am just learning this so I thought it would be good practice.
     
  7. Sep 20, 2004 #6
    Oh sorry I guess tex equations don't turn white
     
  8. Sep 20, 2004 #7

    arildno

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    Besides, you used x as the variable rather than t :wink:
     
  9. Sep 20, 2004 #8
    Darn I was so close to getting it right
     
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