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When acceleration is constant, then distance is a quadratic function of time. So, the solutions can be complex numbers, right?

I can intuitively think of negative solutions like this:

If my scooter is at $x=0$ at $t=0$, it's the initial velocity is $-1ms^{-1}$ and it has a constant acceleration of $2ms^{-2}$, then the my displacement as a function of time is $x=t^2-t$. For, $x=2$, $t=2,-1$. I think of the negative solution, i.e. $-1s$ like this: If, when I started my scooter, time suddenly started to run backwards, then I'd end up at $2m$ after $1s$. So, negative solutions just answer the question, 'what would have happened $1s$ before?'.

But if this is the situation: My scooter is at at rest at $x=0$ at $t=0$, and I start it with a constant acceleration $2ms^{-2}$, then at what time is my displacement $-1m$? Clearly, I can never end up at $-1m$ at any instant, even if time starts running backwards, because $x=t^2$ only takes positive values even for negative $t$. The equation gives the value $t=i$. Can we think of this imaginary time like this?: If when I start my scooter, time suddenly started to run in a perpendicular direction, then I'll end up at $-1m$ after $1s$.

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    $\begingroup$ I'm not sure it has any meaning in terms of simple kinematics, but you may want to check out physics.stackexchange.com/q/46798/45613 $\endgroup$ – docscience Apr 10 '17 at 15:16
  • $\begingroup$ @docscience I don't know all that advanced Physics. Is there no meaning of the complex solutions of these kinematic equations? $\endgroup$ – Dove Apr 10 '17 at 15:23
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    $\begingroup$ Time is real, so an imaginary solution to the equations, means that there is no physical solution. $\endgroup$ – user126422 Apr 10 '17 at 16:59
  • $\begingroup$ What do you mean by time running in a perpendicular direction? Time running backwards is simply asking "What happened before we started the clock ticking (t=0)?" - assuming that the acceleration was still $2m/s^2$. But what is the physical significance of time running in a perpendicular direction? $\endgroup$ – sammy gerbil Apr 12 '17 at 12:43
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    $\begingroup$ I cannot provide an intuitive understanding of time flowing perpendicularly, because I do not know what it means. It is something you suggested, so I expected you to know what it means. So that's two things you want an intuitive explanation of - imaginary time, and time flowing perpendicularly. $\endgroup$ – sammy gerbil Apr 13 '17 at 3:16
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Your logic is flawed. If you plot the function $x = t^2$ on a graph you get a parabola with its minimum at $t = 0$ and a slope, $\frac{dx}{dt} = 0$ at that point. The negative values of $t$ give positive values for $x$ and indicate what the object was doing before you started your clock. It came in from $+x$, slowed to a stop at $t = 0$, and then accelerated back toward $+x$. The results are similar for $x = t^2 – t$, but the minimum occurs at a $+t$ and $–x$. If you don't get a real time for a particular position, it just means that the object being described by your equation does not pass through that position.

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The only thing that I can currently think of if time was a complex quantity is that it wouldn't have any forward direction, which isn't so in our macroscopic case. Here if you drop an egg then you can always denote it's position with respect to time in an ever increasing order of time where the egg never gets back to it's original state.

But it does have use in Quantum Mechanics, etc. you can read about it over here

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I am not sure how you got $x = 1 \mathrm m$ at $t = i$, but it is not true. At $t = i$, $x = [-1 + -i]$, not $i$ !

Keep in mind that the given equation places a constraint on the possible values $x$ can have for a given value of $t$. Also, if you allow $t$ to be a complex number, then you no longer are dealing with a plane $(x, t)$, but rather a volume $(x, t, i)$. If this is the case, then the real part of $t$ will be on the $t$ axis and the imaginary part will be on the $i$ axis. Since $i$ is perpendicular to $x$ and t axes, then it doesn't matter how large $t$ and $i$ become, the vector formed, will never intersect the $x$ axis other than at $(0,0,0)$.

By the way, the minimum value for $x$, is obtained at $t = \dfrac12$, with $x = -\dfrac14$.

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    $\begingroup$ No, in the second situation, I've assumed my scooter to be initially at rest, so $t=i, -i$. And, why is the intersection with $x-axis$ so important? $\endgroup$ – Dove Apr 13 '17 at 1:51
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Here is an extract from "The Universe In a Nutshell"

One might think this means that imaginary numbers are just a mathematical game having nothing to do with the real world. From the viewpoint of positivist philosophy, however, one cannot determine what is real. All one can do is find which mathematical models describe the universe we live in. It turns out that a mathematical model involving imaginary time predicts not only effects we have already observed but also effects we have not been able to measure yet nevertheless believe in for other reasons. So what is real and what is imaginary? Is the distinction just in our minds?

Imaginary time is real time which has undergone a Wick rotation so that its coordinates are multiplied by $i$. Imaginary time is not imaginary in the sense that it is unreal or made-up (any more than, say, irrational numbers defy logic), it is simply expressed in terms of what mathematicians call imaginary numbers.

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  • $\begingroup$ I read that book. It's full of pseudoscience and misinterpretations from someone that was supposed to know the science well. I wouldn't trust any interpretation from it. Doesn't it also say that an electron's spin 1/2 means an electron has to rotate around 4pi radians for the same side to be facing you? Yeah, don't buy that stuff $\endgroup$ – Jim Dec 20 '19 at 20:56

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