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Math Help - Integrating velocity

  1. #1
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    Integrating velocity

    Hi,

    I'm trying to calculate the position of an object that moves ahead with constant acceleration but that changes its heading at a constant rate. I get a strange result, so I think that I'm missing something. Could you take a look and give me some advice?

    The linear velocity of the object is v = v_0 + a t, and the heading is \theta = \theta_0 + \omega t. The velocity in x is then v_x = (v_0 + a t) \cos(\theta_0 + \omega t).

    To simplify, I consider the case with v_0=0, \theta_0=0, a=1, so v_x = t \cos \omega t.

    Then I try to integrate v_x to get the position in x, and I get the following:

    x(t) = \frac{\cos \omega t}{\omega^2} + \frac{t \sin \omega t}{\omega}

    My problem is that if now I consider the case where the object moves with constant heading, \omega = 0, and I apply the formula above, the result is infinity.

    What I am doing wrong here?

    Thanks!
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  2. #2
    MHF Contributor chisigma's Avatar
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    The value of x(t) is...

    x(t)= x_{0} + \int_{0}^{t} v_{x}(\xi)\cdot d\xi = x_{0} + \int_{0}^{t} \xi \cdot \cos \omega \xi \cdot d\xi= x_{0} + \frac{t\cdot \sin \omega t}{\omega} - \frac{1-\cos \omega t}{\omega^{2}} (1)

    Now if you compute the \lim_{\omega \rightarrow 0} of the (1), you obtain a finite value...

    Kind regards

    \chi \sigma
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  3. #3
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    You're assuming that your w is non-zero when you integrate. If you have w=0 then your function ceases to be t * cos(wt) and turns into t.

    Since you're integrating with respect to t, the 'w' value is important as it defines what kind of function you're integrating. That is, when you integrate cos(wt) if you have w=0, you don't have cos(t) anymore.

    In that case you should just integrate t. By the way, you're missing a constant when you integrate.
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  4. #4
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    Quote Originally Posted by chisigma View Post
    The value of x(t) is...

    x(t)= x_{0} + \int_{0}^{t} v_{x}(\xi)\cdot d\xi = x_{0} + \int_{0}^{t} \xi \cdot \cos \omega \xi \cdot d\xi= x_{0} + \frac{t\cdot \sin \omega t}{\omega} - \frac{1-\cos \omega t}{\omega^{2}} (1)

    Now if you compute the \lim_{\omega \rightarrow 0} of the (1), you obtain a finite value...

    Kind regards

    \chi \sigma
    That's right
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  5. #5
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    Thanks a lot!!!

    Integrate between 0 and t. Now, that makes sense...
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