$$s(t) = 4t + \frac2{t-3} + \frac23$$

where $s$ represents the position of either particle in the tubular cloud-chamber measured in centimetres, while $t$ represents the time in nano-seconds.

It's for the following problem:

A sub-atomic particle is travelling in a straight line through a tubular cloud-chamber that is 28 cm long. The particle is subjected to an electromagnetic field that reverses the direction of the particle so that it disappears from the cloud-chamber. Almost instantaneously with the disappearance of the first particle, a second particle enters the chamber from the other direction. This particle is subjected to the same electromagnetic field so that its direction is also reversed and it disappears from the cloud-chamber after a period of time along the same trajectory as the first particle would have done if the electromagnetic field had not been applied.

  • $\begingroup$ Does $2/t-3$ mean $(2/t)-3$ or $2/(t-3)$? $\endgroup$ – Gerry Myerson Mar 2 '12 at 1:32
  • $\begingroup$ The latter: \begin{equation}(2/(t-3))\end{equation} $\endgroup$ – Christy McGrory Mar 2 '12 at 1:34

Try multiplying both sides by $3(t-3)$, bring all terms to one side and then use the quadratic formula.

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  • $\begingroup$ For some reason my method is way off... I'm not able to get to a point where I can use the quadratic. $\endgroup$ – Christy McGrory Mar 2 '12 at 2:09
  • 1
    $\begingroup$ Well, what do you get when you do as Emile says and multiply both sides by $3(t-3)$ (and do such cancellations as are available)? $\endgroup$ – Gerry Myerson Mar 2 '12 at 2:55
  • $\begingroup$ I get \begin{equation} 4t^2 = 3s + st - 3s + 2 \end{equation} but then obviously I can't go anywhere from there. Any help with my method would be appreciated. $\endgroup$ – Christy McGrory Mar 2 '12 at 3:06
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    $\begingroup$ @ChristyMcGrory: Obviously! Oh, wait, maybe you can do something. But first you should check your algebra. Multiplying both sides of $$s = 4t + \frac2{t-3} + \frac23$$ by $3(t-3)$ you get $$3s(t-3) = 12t(t-3) + 6 + 2(t-3)$$ which rearranges to $$12t^2 - (34+3s)t + 9s = 0.$$ Then just use the quadratic formula to solve this equation for $t$ (let $a = 12$, $b = -34-3s$, and $c = 9s$ in this formula). $\endgroup$ – Antonio Vargas Mar 2 '12 at 5:17

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