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Calculating instantaneous acceleration

  1. Jul 30, 2013 #1
    1. The problem statement, all variables and given/known data

    The position of a particle moving along the x axis is given in centimeters by x = 9.75 + 1.50t3, where t is in seconds

    Calculate the instantaneous acceleration at 2.00 s.

    2. Relevant equations
    x = 9.75 + 1.50t3

    I don't understand how i would solve for the acceleration using an equation for distance
    i understand Vf=Vo+at and that the instantaneous velocity at this time is .217 m/s


    i had to teach myself how to do this so i am not sure i found the velocity correctly
     
    Last edited: Jul 30, 2013
  2. jcsd
  3. Jul 30, 2013 #2
    The way this works in PF is that you are supposed to make an attempt to solve the problem first, and show us what you did so far. Then, if necessary, we can try to get you pointed in the right direction.

    Chet
     
  4. Jul 30, 2013 #3
    Thanks for the info
     
  5. Jul 30, 2013 #4
    You need to derive it. By t subject, do it twice. One derive tell you velocity, second tell you acceleration.
     
  6. Jul 30, 2013 #5
    How did you arrive at v(2.00s) = 0.217 m/s? Are you familiar with what the derivative of position means?
     
  7. Jul 30, 2013 #6
    how would i derive it the second time
     
  8. Jul 30, 2013 #7
    You have a function for position with respect to time. Take the first time derivative. Show us what you get and what does this new function represent?
     
  9. Jul 30, 2013 #8
    Did u ask it from me? I cant understand, sorry my bad english. Am i wrong about the derivation?
     
  10. Jul 30, 2013 #9
    No, sorry, that was directed at maxalador.
     
  11. Jul 30, 2013 #10
    Oh okay. :)
     
  12. Jul 30, 2013 #11
    sorry guys it turns out that the reason i didnt understand was because i was doing it wrong. thanks for all your help though
     
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