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Very Simple Integral ( a dt )

  1. Mar 25, 2012 #1
    Edit: acceleration is not constant.

    1. The problem statement, all variables and given/known data

    [tex] v = ∫ a dt [/tex]

    2. Relevant equations



    3. The attempt at a solution

    I tried integration by parts but I'm clearly lacking the conceptual mind-set for this problem. I guess I got too used to mindlessly doing integration problems and when it came to something a bit different I got stumped.

    Am I too complicating things by using integration by parts or is it necessary?? If it is necessary then I will post my step-by-step solution. But I'm not sure its right-- I plugged the problem in Wolframalpha and it spit out [tex] v = at [/tex]
     
    Last edited: Mar 25, 2012
  2. jcsd
  3. Mar 25, 2012 #2
    If a is a constant, then you really have:

    [tex]v = a\int 1 dt[/tex]
    What's the integral of a constant, in this case, 1?
     
  4. Mar 25, 2012 #3
    do you know how to integrate ∫dt ?
     
  5. Mar 25, 2012 #4

    jhae2.718

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    This appears to be an integral relating acceleration and velocity. Acceleration is the derivative of velocity with respect to time, [itex]a=\frac{{\rm d}v}{{\rm d}t}[/itex]. If we rearrange this, we get:
    [tex]
    v = \int\!a{\rm d}t
    [/tex]
    If [itex]a[/itex] is constant, this integral evaluates to [itex]v(t)=at+v_0[/itex].

    If [itex]a=a(t)[/itex] is a function of time, we need to know more about [itex]a(t)[/itex] to evaluate the integral.
     
    Last edited: Mar 25, 2012
  6. Mar 25, 2012 #5

    rock.freak667

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    What are you defining 'a' as ? It looks like acceleration to me, but from Wolfram's perspective 'a' is just a constant, so the result would be 'at'.
     
  7. Mar 25, 2012 #6
    Sorry, a is not constant. This has to do with the acceleration of a gravitational field (I'm going to generalize it later).
     
    Last edited: Mar 25, 2012
  8. Mar 25, 2012 #7
    What do you need to know about a(t)? I'm just trying to find a general expression for velocity under varying acceleration.
     
    Last edited: Mar 25, 2012
  9. Mar 25, 2012 #8

    jhae2.718

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    You already have that:
    [tex]
    {\rm d}v = a(t){\rm d}t
    [/tex]
     
  10. Mar 25, 2012 #9
    Then [tex] v = ∫ a dt [/tex]

    But I'm not following what your trying to say exactly?
     
  11. Mar 25, 2012 #10

    jhae2.718

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    What I'm saying is that that's as general as you're going to get. You'll need to evaluate the expression for whatever specific [itex]a(t)[/itex] you have to get a meaningful kinematic expression.
     
  12. Mar 25, 2012 #11
    I have trouble believing that. What about gravitational fields where acceleration isn't the same?
     
  13. Mar 25, 2012 #12
    If it irks you, how about you help me integrate the equivalent:

    [tex] a = \frac{ g m_1}{r^2} [/tex]
    [tex] {G m_1}∫ r^{-2} dt [/tex]
     
  14. Mar 25, 2012 #13

    rock.freak667

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    What jhae2.718 was saying is that you were trying to integrate 'a' without knowing the function.

    In general v= ∫a dt. Without knowing what function 'a' is, you can't say what 'v' will be.

    If you said a= 2t, then you could say v=∫2t dt = t2+C
     
  15. Mar 25, 2012 #14

    jhae2.718

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    Without knowing what [itex]r[/itex] is as a function of time, you can't integrate that. If you leave it in differential form, you can get the second order nonlinear ordinary differential equation below, which can be solved to find [itex]r(t)[/itex] and then differentiated to find [itex]v(t)[/itex]:
    [tex]\ddot{r} - \frac{Gm_1}{r^2} = 0[/tex]
    Avoiding solving this ODE, you can use the chain rule to get an expression relating [itex]r[/itex] and [itex]\frac{{\rm d}a}{{\rm d}r}[/itex], which you can integrate. (The resulting expression you will find to be conservation of mechanical energy.)
     
  16. Mar 25, 2012 #15
    So that is why I couldn't solve it.. Second order nonlinear ODE. I'm only in Calculus II right now. I've been looking all over the Calculus book for help but now I see why I didn't find anything :rolleyes: .

    Oh and by the way, I don't have a clue on how to use the chain rule on this expression. :confused:
     
  17. Mar 25, 2012 #16
    I'm just throwing a guess in here, but if your goal is to find the velocity of something which is under the influence of a non-constant gravitational field, integrating its acceleration is not the easiest way to go. You would typically apply conservation of energy between initial and final stages:
    [tex]K_i + U_i = K_f + U_f[/tex]
    where [itex]K=mv^2/2[/itex] represents kinetic energy and [itex]U=-GMm/r[/itex] represents potential energy.
     
  18. Mar 25, 2012 #17
    No I'm just doing an independent "research project", just for fun. At this beginning stage, I'm trying to describe the gravitational field in terms of a varying field of kinetic energy with test mass [itex]m_o[/itex].

    Which is why I was trying to find the velocity under non-constant acceleration. I have a lot of work to do because it seems that I need to know some non-linear ODE whilst I'm only in Calculus II.
     
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