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Possible Duplicate:
How would I rearrange this equation to make the subject $t$?

The scenario is as follows:

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. It is possible to model approximately these events with the function

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

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

The question is:

When does the first particle actually leave the cloud-chamber according to the model?

[Hint: consider s(t) = 0]

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marked as duplicate by Gerry Myerson, Kannappan Sampath, Henning Makholm, Asaf Karagila, Zev Chonoles Mar 4 '12 at 9:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

That's some pretty weird accelerations to result from an electomagnetic field ... – Henning Makholm Mar 3 '12 at 3:05
In facts, it looks like at $t=2.8$ the first particle is still inside the chamber, but moving significantly faster than the speed of light. – Henning Makholm Mar 3 '12 at 3:12
So, what didn't you understand about the solution to this problem when you posted it as…? – Gerry Myerson Mar 3 '12 at 3:33
If? OK, I'll spell it out for you. You are asking people for help. You owe it to them to give them all the relevant information. Having already asked, and having been given answers, to (pretty much) the same question is bloody well relevant information. Etiquette demands that you link to the other question. Even better would be to just edit the other question to include what you are asking here, rather than posting a new question. And best of all would be working out how you could use what you were told about the other question to figure out the one you have now. – Gerry Myerson Mar 3 '12 at 4:24
@Christy McGrory: You are welcome. You had carried out the multiplications successfully, and had more or less arrived at $12t^2=34t=0$. It would have been overkill (but possible) to use the Quadratic Formula on that, since $12t^2-34t$ factors so simply. – André Nicolas Mar 3 '12 at 5:05
up vote 3 down vote accepted

You were given the hint (or command?) to find the time $t$ when $s(t)=0$. We will do that, although the relevance to the problem as stated is not entirely clear. So we want to solve the equation $$4t + \frac{2}{t-3} + \frac{2}{3}=0.$$ Multiply through by $3(t-3)$. This is legitimate, since we cannot have $t=3$. We arrive at the equivalent equation $$(3)(t-3)(4t)+(3)(2)+(t-3)(2)=0.$$ This simplifies to $$12t^2-34t=0, \quad\text{and then to}\quad t(12t-34)=0.$$ Since the solution $t=0$ is irrelevant. we arrive at the equation $12t-34=0$, whose solution is $t=\dfrac{34}{12}$.

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