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If acceleration is stated is terms of velocity...

  1. Dec 8, 2016 #1
    1. The problem statement, all variables and given/known data
    This is not a particular problem but a generic one. If one has acceleration stated in terms of velocity (or position) such as a=v, how do we convert these values into time?

    2. Relevant equations
    ads=vdv
    ad(theta)=wdw

    And of course, the usual time-based derivative relations for acceleration, velocity and positions.

    3. The attempt at a solution
    Sorry, guys, this one is killing me right now.
     
  2. jcsd
  3. Dec 8, 2016 #2

    Orodruin

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    First of all, this cannot be the relationship as it is dimensionally inconsistent.

    Second, what you have is just a differential equation that you need to solve.
     
  4. Dec 8, 2016 #3

    cnh1995

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    You can use the integral-differential relationship between the two quantities. Also, a=v is dimensionally incorrect. The equation could be a=kv, where k has the unit of time-1.
     
  5. Dec 11, 2016 #4
    Allright, clarified question... let's saw we have $$\alpha=50\sqrt{\omega}$$. This is, I believe, equivalent to the diffyQ $$y''=50\sqrt{y'}$$. The book solves the equation by taking from $$dw/dt=a $$ $$ dt=dw/a$$, and integrating with bounds that are suggested by the problem. Wolframalpha solves the problem by giving $$y=\frac{625t^3}{3}+c_1^2+25c_1t^2+c_2$$

    So my real question is.. how do I reconcile these two, or am I wrong about something? (I can't find appropriate values of the constants of integration in the case of the last equation!)
     
  6. Dec 11, 2016 #5
    In that case, ##y## is ##\theta## and the rest is the same, although I think Wolfram Alpha must have had a factor of ##t## on the ##c_1^2##, did it not?

    Since ##\alpha = d\omega / dt##, you can rewrite the first equation as:

    $$\frac{d\omega}{ dt} = 50\sqrt{\omega}$$

    And then through separation of variables:

    $$\frac{d\omega}{50\sqrt{\omega}} = dt$$

    Then integrate both sides:

    $$\frac{1}{50}\int\frac{d\omega}{\sqrt{\omega}} = \int dt$$

    $$\frac{1}{25}\sqrt{\omega} = t + c_1$$

    $$\omega= 625t^2 + 1250c_1t + 625c_1^2$$

    Note that if we roll a factor of 25 into ##c_1##, this becomes:

    $$\omega= 625t^2 + 50c_1t + c_1^2$$

    This is ok to do because the constants are arbitrary. And since ##\omega = d\theta / dt##, you can repeat the process:

    $$\frac{d\theta}{dt} = 625t^2 + 50c_1t + c_1^2$$

    $$d\theta = (625t^2 + 50c_1t + c_1^2) dt$$

    $$\int d\theta = \int (625t^2 + 50c_1t + c_1^2) dt$$

    $$\theta = \frac{625}{3}t^3 + 25c_1t^2 + c_1^2 t + c_2$$

    As for the boundary conditions, you can redo the above using the boundary conditions as the limits of integration. Or, you can use the above equation to plug the values for ##y## and ##t## that you are given, and solve for the arbitrary constants. You'll need two conditions to solve for both constants.
     
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