# If my acceleration is -1 ($a=-1\:\rm{m/s^2}$) and I'm standing in the infinite ($x_0=\infty \:\rm m$), could I reach the point $x=0\:\rm m$?

I'm standing in the infinite where $x_0=\infty \:\rm m$. If I have a negative acceleration, could I reach the point $x=0\:\rm m$? Would it be possible to calculate how long would take to reach the point $x=0\:\rm m$?

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No you won't reach $x=0\,\rm m$. – ja72 Mar 23 '12 at 23:43
You can't really talk about 'standing at' infinity. In physics when we talk about taking a parameter to infinity we really mean "too big for us to care about its actual value". So for example, if we say the heat capacity of a gas "at $T=\infty$" is constant, we just mean the temperature is too high for us to see quantum effects - you can't have infinite temperature in the real world. – Benjamin Hodgson Aug 16 '12 at 22:46
If you're standing at $x_0=\infty\textrm{ m}$ and move $1\textrm{ m}$ to the left, where do you end up? – Emilio Pisanty Aug 17 '12 at 0:41

(Mathematicians, please hold back the pitchforks--i'll update this with a mathematically correct version later.)

$$s=ut+\frac{1}{2}at^2$$,$$\text{where }u=0,a=-1,s=-\infty$$

$$\therefore-\infty=0+\frac{1}{2}(-1)t^2$$ $$\therefore t=\sqrt{2\infty}=\infty \:^\text{ {*}}$$

So you will never reach $s=0$. Of course, standing at $s=\infty$ is impossible anyways. In physics, infinity is a place sufficiently removed from a system to be free of all influences from it. For example, a hundred meters is infinity for a system consisting of a marble, where you only consider gravity. So it's a relative concept. In mathematics, infinity is something that you can try to reach but never do--it's not a real number(in both the lay and technical meanings of "real number").

A more mathematically correct way to do it would be to use limits and show that $t$ diverges as $s$ approaches $-\infty$.

*Note that even though I am using it like a normal number, $\infty$ is not a number and arithmetic does not work on it. $\infty+\infty=\infty\times\infty=\infty^\infty=\infty$ is OK to use, but $\infty-\infty,\infty\times0,\infty/\infty$ are all undefined like $0/0$ and should not be used.

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You don't have to argue in a formal system--- this answer is good enough. You can just replace $\infty$ with a large number and it is manifest that the time diverges. – Ron Maimon Mar 24 '12 at 7:06
@RonMaimon True--but I always cringe whenever I see arithmetic being done with $\infty$--it's an abuse of notation since it appears as if $\infty$ is a number which you can subtract. The correct way is to use limits and get that it diverges like you said, but I wasn't sure if the OP would understand that.. – Manishearth Mar 24 '12 at 7:46
Footnoted it and added a bit on limits. – Manishearth Mar 24 '12 at 7:54

It's same as you start from $x_0=0\:\rm m$ and have $a=1\:\rm{m/s^2}$. You will aim to $\infty$, but never reach it.

Sorry, didn't find how to write with math code.

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Welcome to physics.stackexchange!! To use math, enclose the TeX code in dollar signs: $.....$. Using double dollar signs will make the math larger and on its own line. See physics.stackexchange.com/faq#notation . if you want, I can get you some links on learning simple math TeX. – Manishearth Mar 24 '12 at 7:58
Thanks, this link helped a lot. – Narek Mar 24 '12 at 8:02
Tutorial sort of thing: en.wikibooks.org/wiki/LaTeX/Mathematics ; good reference: enwp.org/WP:MATH – Manishearth Mar 24 '12 at 8:05

The given answers are not really correct. Manishearth should have followed his intuition. Maybe this might help: $$x_0 = \infty \\x(t) = x_0 +\frac{1}{2}at^2$$ Now can we reach a certain point after waiting for a long time? $$\lim_{t\rightarrow \infty}x(t) = x_0 +\frac{1}{2}at^2= x_0 - \infty \\ \lim_{t\rightarrow \infty}x(t) = \infty - \infty$$ So now where are stuck and ask our math guys what is $\infty - \infty$? and it turns out that we just forgot how we came to our starting position. If we do not know that (i.e. which limit brought us there), then the answer is not defined as just the symbol $\infty$ has no memory. So yes, we could certainly be at $0$, but also at $-\infty$ or$\infty$. A precise answer can only be given if we know how we got there.

This is also true for the time. If you do not know how far you are out you cannot calculate how long it will take to come back.

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This seems a little disingenuous. You are trying to take two different limits in the last line ($x_0 \to \infty$ and $t \to \infty$), which overconstrains your problem. The correct way to do it is to fix $x(t)=0$ and take one of the two limits, as Manishearth has (non-rigourously) done. – Benjamin Hodgson Aug 16 '12 at 22:41
@poorsod: I was not trying to be disingenuous. Isn't this the the whole point that there are two limits and the result is indeterminate? So unless we have additional information the answer is that we can end up anywhere, including $0$. – Alexander Aug 17 '12 at 2:49
There is only one limit: $t$ diverges in the limit $x_0 \to \infty$. We are taking the limit of $x_0$ and observing $t$'s behaviour. – Benjamin Hodgson Aug 18 '12 at 20:26
@poorsod: Under that condition to reach $x=0$ we need the time $t=\sqrt(-2 x_0/a)$, so not only can we reach it but we can also calculate the time needed. So why should we not say yes to answer the original question? – Alexander Aug 18 '12 at 21:27
Well, the original question is slightly ill-posed, since you can't 'stand at' infinity (it's not a place!). However, for every $x_0 < \infty$, $t$ does have a solution, as you pointed out. It diverges as $x_0\to\infty$. So I would say the answer to the original question is indeed yes, but with certain qualifications. – Benjamin Hodgson Aug 19 '12 at 0:07