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Finding the initial velocity, please help.

  1. Sep 2, 2011 #1
    1. The problem statement, all variables and given/known data
    The position x of an object as a function of time is given as x = 2.257 + 4.894 t + 16.5 t2. All quantities are understood to be quoted in SI units. What is the initial velocity ? Indicate with a negative (positive) sign whether the acceleration is constant (not constant) for the given expression.


    2. Relevant equations
    i think i have to use. x=x0+v0(t)+1/2(a)t^2)


    3. The attempt at a solution
    the problem is confusing me becuase im not sure what numbers should go where in the equation. i know the initial velocity is v0.
     
  2. jcsd
  3. Sep 2, 2011 #2

    olivermsun

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    So can you compare the equation you wrote in (2) with the one you wrote in (1)?
     
  4. Sep 2, 2011 #3
    2.257 would be the x0, 4.894 would be the t, and 16.5 would go where the t^2 is. im not sure what to put as the a.
     
  5. Sep 2, 2011 #4

    tiny-tim

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    welcome to pf!

    hi megkirch! welcome to pf! :wink:
    yes :smile:
    16.5 is the ½a :wink:
     
  6. Sep 2, 2011 #5
    ohh ok thank you! So it would be, x=2.257+V0(4.894)+1/2(16.5)?
     
  7. Sep 2, 2011 #6

    tiny-tim

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    i hesitate to dampen your enthusiasm, but that doesn't really make any sense :tongue2:

    would you like to rephrase it? :smile:
     
  8. Sep 2, 2011 #7
    do i have that equation set up right?
     
  9. Sep 2, 2011 #8
    i have been stuck on this one problem all day, please help.
     
  10. Sep 2, 2011 #9
    What I would do is derive the equation using nx^n-1 where n represents the power in terms of time and you will get the equation that represents velocity.

    from this equation, v= v0 + at, what can you say about v0 in the equation that you derived. (hint: the derived equation is identical to the velocity equation)
     
  11. Sep 2, 2011 #10

    tiny-tim

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    the equation is x = 2.257 + 4.894 t + 16.5 t2

    you have to find v0 and a …

    so what is a?​
     
  12. Sep 2, 2011 #11
    sorry, im completely lost
     
  13. Sep 2, 2011 #12
    do you know how to find a derivative?
     
  14. Sep 2, 2011 #13
    yes the derivative of the equations is. 2.257=0 4.894t=4.894 16.5t^2=33.
     
  15. Sep 2, 2011 #14
    thank you so much! i finally got the answer its 4.894
     
  16. Sep 2, 2011 #15
    thats what I think :) want a second opinion? I mean I'm sure that's right.
     
  17. Sep 2, 2011 #16
    my hw is online it said its right. thank you again
     
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