# Integrating velocity

• Jun 26th 2009, 02:09 AM
mgma
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!
• Jun 26th 2009, 03:07 AM
chisigma
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

$\chi$ $\sigma$
• Jun 26th 2009, 03:09 AM
pedrosorio
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.
• Jun 26th 2009, 03:11 AM
pedrosorio
Quote:

Originally Posted by chisigma
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

$\chi$ $\sigma$