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Simple dynamic accell problem

  1. Sep 6, 2007 #1
    The acceleration of a particle along an x axis is a = 5.0t, with t in seconds and a in meters per second squared. At t = 2.0 s, its velocity is +13 m/s. What is its velocity at t = 6.5 s?




    I dont really know how to go about this, as it is a dynamic acceleration, any help would be greatly appreciated

    so far all I have got is a=deltaV/deltaT
     
  2. jcsd
  3. Sep 6, 2007 #2

    Doc Al

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    Staff: Mentor

    That's OK for average acceleration, but for instantaneous acceleration use a derivative: [itex]a = dv/dt[/itex].

    Given acceleration, how do you find velocity? Hint: Use a little calculus.
     
  4. Sep 6, 2007 #3
    well I would multiply by the time passage, but again this only works for constant acell right? my problem is the derivative, I know how to do them, but i don't understand how that applies. the derivative of a is 5 right? thats great, but what do I do after that
     
  5. Sep 6, 2007 #4

    Doc Al

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    If you were given v, you'd take its derivative to find a. But you have the opposite situation: You are given a and need to find v. What do you do?
     
  6. Sep 6, 2007 #5
    integrate? so 2.5t^2=v
     
  7. Sep 6, 2007 #6

    Doc Al

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    Good! But don't forget the constant of integration.
     
  8. Sep 6, 2007 #7
    wow awesome thanks a lot.
     
  9. Jan 12, 2008 #8
    Finding position

    I have run across the same type of problem, but I am being asked what the position is. Below is what was given.

    Acceleration of a point is a=20t m/s^2. When t=0, s=40 meters and v=-10 m/s. What is the position at t=3 sec?


    I've used a=dv/dt to get the velocity, v=80 m/s.

    I'm not sure were to start to get the position. Can you advise?

    Thank you.
     
  10. Jan 13, 2008 #9

    Doc Al

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    Using a=dv/dt, you integrated to get the velocity. But v = dx/dt, so integrate once again to get the position.
     
  11. Jan 13, 2008 #10
    For velocity I got v=10t^2 + c, and pluged 3 in for t and -10 in for c. This gave me v=80 m/s.

    If i integrate once more to get position I get s=(10t^3)/3+ct.

    Am I correct with the constant?
    Plug 3 in for t, and -10 in for c again.

    I think Im getting confused with the constant.
     
  12. Jan 13, 2008 #11

    Doc Al

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    That's the velocity at t = 3; in general it's: v=10t^2 - 10.

    But you forgot the new constant of integration. (Hint: You are given the position at t = 0.)
     
  13. Jan 13, 2008 #12
    Thats what I was forgetting.

    Therefore s=(10t^3)/3+ct+d.

    Where
    t=3
    c=-10
    d=40

    Thanks for the help.
     
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