Davorak
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Just read Bartholomew's post oh well. Chronon did somthing like this:
<br /> \int \frac{1}{{\sqrt{c - a\,x^2}}} \,dx = \frac{\arcsin ({\sqrt{\frac{a}{c}}}\,x)}{{\sqrt{\Mfunction{a}}}}<br />
substitute in c\ for \ V^2 \ and \ a \ for \ \frac{r^2\,V^2}{R^2}
so using substituting in we get
<br /> \int \frac{1}{{\sqrt{V^2 - \frac{r^2\,V^2}{R^2}}}}\,dr = \frac{\arcsin (r\,{\sqrt{R^{-2}}})}{{\sqrt{\frac{{\Mfunction{V}}^2}{R^2}}}} = \frac{R\,\arcsin (r\,{\sqrt{R^{-2}}})}{V} = t<br />
Therefore
<br /> r = R\,\sin (\frac{t\,V}{R})<br />
<br /> r = R \ When \ \sin (\frac{t\,V}{R})=1<br />
Therefore
<br /> \frac{t\,V}{R} = \frac{\pi }{2}<br />
<br /> t = \frac{\pi \,R}{2\,V}<br />
I did not include \pm for square root or multiple solutions but in the end they do not matter.
I still like my idea of replacing with a force and using conservation of energy to show the cat will catch the mouse. If anyone would like to see it let me know.
<br /> \int \frac{1}{{\sqrt{c - a\,x^2}}} \,dx = \frac{\arcsin ({\sqrt{\frac{a}{c}}}\,x)}{{\sqrt{\Mfunction{a}}}}<br />
substitute in c\ for \ V^2 \ and \ a \ for \ \frac{r^2\,V^2}{R^2}
so using substituting in we get
<br /> \int \frac{1}{{\sqrt{V^2 - \frac{r^2\,V^2}{R^2}}}}\,dr = \frac{\arcsin (r\,{\sqrt{R^{-2}}})}{{\sqrt{\frac{{\Mfunction{V}}^2}{R^2}}}} = \frac{R\,\arcsin (r\,{\sqrt{R^{-2}}})}{V} = t<br />
Therefore
<br /> r = R\,\sin (\frac{t\,V}{R})<br />
<br /> r = R \ When \ \sin (\frac{t\,V}{R})=1<br />
Therefore
<br /> \frac{t\,V}{R} = \frac{\pi }{2}<br />
<br /> t = \frac{\pi \,R}{2\,V}<br />
I did not include \pm for square root or multiple solutions but in the end they do not matter.
I still like my idea of replacing with a force and using conservation of energy to show the cat will catch the mouse. If anyone would like to see it let me know.