# Displacement of a particle

1. Jul 15, 2014

### FilupSmith

A few months ago, I stumped my Mathematics teacher with a question when we were learning about displacement of a particle, given a formula. For example, $x=t^{2}-t-1$, where x is in meters and t is in seconds.

Anyway, she made it very clear how to solve displacement when given time t (Simply solving for the unknown value). My question was, what happens in the case of, for example, the displacement $x=\sqrt{t-2}$, when t=1 second?

I guess my question is, how is displacement effected if the displacement x is a complex value?

2. Jul 15, 2014

### HakimPhilo

Since you're working on real valued functions, you must restrict the domain of the function to values that produce real numbers. In other words, the displacement is undefined when your function produces complex values.

Last edited: Jul 15, 2014
3. Jul 15, 2014

### FilupSmith

That sounds most probable.
So for the case of $x=\sqrt{t-2}$, $x\in \mathbb{R}$ ?

~| FilupSmith |~

4. Jul 15, 2014

You would like to see the graph of the equation $y=\sqrt{x-2}$

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5. Jul 15, 2014

### HakimPhilo

The variable here is $t$, for the function to be defined what's inside the square root must be positive, i.e. $t-2\geqslant0\Rightarrow t\geqslant 2$.

Last edited: Jul 15, 2014
6. Jul 15, 2014

### FilupSmith

Ah, I see.

~| FilupSmith |~

7. Jul 15, 2014

### FilupSmith

I know what it looks like, I'm just curious about what the displacement would be for non-real x values - but it seems to be that for $x=f\left( t\right)$, x cannot be complex.

~| FilupSmith |~