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Easiest problem that I can't figure out

  1. Feb 18, 2013 #1
    say your are giving the acceleration as a=2x and you know it has an initial velocity of 5m/s at x=0,t=0
    How the heck do you find the position as function of time??
    It seems so simple but really has me stumped. It must be possible.

    I have played around with vdv=ads and integrated trying to unlock time..
     
  2. jcsd
  3. Feb 18, 2013 #2

    SammyS

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    Write [itex]\ a=2x\ [/itex] as [itex]\displaystyle\ \frac{dv}{dt}=2x\ .[/itex]

    Then use the chain rule to write [itex]\displaystyle\ \frac{dv}{dt}\ [/itex] as [itex]\displaystyle\ \frac{dv}{dx}\,\frac{dx}{dt}\,,\ [/itex] but realize that [itex]\displaystyle\ \frac{dx}{dt}=v\ .[/itex]

    That leaves you with a separable differential equation with variables, x and v .
     
  4. Feb 18, 2013 #3
    Ya I agree, and once you separate variables and integrate you can obtain v=sqrt(2*x^2+25)
    That leaves me no closer to finding the position as function of time...
     
    Last edited: Feb 18, 2013
  5. Feb 18, 2013 #4

    SammyS

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    That's not quite right.

    It gives v = √(10) when x = 0 .

    Once you fix that, remember, [itex]\displaystyle\ v=\frac{dx}{dt}\ .[/itex]
     
  6. Feb 18, 2013 #5
    My bad I meant to write 25. My initial question was asking for the position as a function of time as the particle moves down the positive x-axis given the specific initial conditions stated above.
    I don't care about the velocity as a function of x nor the velocity at x=0. Those are all easily obtainable..
     
    Last edited: Feb 18, 2013
  7. Feb 18, 2013 #6
    I am going to move this post to advanced physics. Please don't get mad at me moderators as I think there is a simple solution to this.
     
  8. Feb 18, 2013 #7
    Hardest problem that I can't figure out...

    say your are giving the acceleration as a=2x and you know it has an initial velocity of 5m/s at x=0,t=0
    How the heck do you find the position as function of time??
    It seems so simple but really has me stumped. It must be possible.

    Moderator's Note: Multiple threads merged​
     
    Last edited by a moderator: Feb 19, 2013
  9. Feb 18, 2013 #8

    TSny

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    Re: Hardest problem that I can't figure out...

    The trick is to use the chain rule to write a = dv/dt = dv/dx ##\cdot## dx/dt = v dv/dx.

    See if you can use this to find v as a function of x. Then see if you can figure out a way to determine x as a function of t.
     
  10. Feb 18, 2013 #9

    Pengwuino

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    Re: Hardest problem that I can't figure out...

    Are you familiar with differential equations? Are you sure it's not [itex]a(t) = 2t[/itex]? If you really mean [itex]a(t)= 2x(t)[/itex], then you have a second order differential equation.
     
  11. Feb 18, 2013 #10
    Re: Hardest problem that I can't figure out...

    oh my god I am such an idiot for not recognizing the diffeq... Thank you
    I will put my head in a hole now
     
  12. Feb 18, 2013 #11
    This thread can be deleted
     
  13. Feb 18, 2013 #12

    SammyS

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    As I said, [itex]\displaystyle\ \frac{dx}{dt}=v\ .[/itex]

    That gives you the differential equation:
    [itex]\displaystyle\ \frac{dx}{dt}=\sqrt{2\,x^2+25\,\ .}[/itex]​



    ... and you should care about the velocity at x=0 . That's one of your boundary conditions.
     
  14. Feb 18, 2013 #13

    TSny

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    Re: Hardest problem that I can't figure out...

    Yes, if you are familiar with 2nd order differential equations, Pengwuino's suggestion is the way to go. My suggestion involves solving a couple of first-order differential equations using separation of variables and takes more steps. Not as good.
     
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