\(\displaystyle t\) does not really have a physical meaning. You can sort of think of it as a dummy variable.

The reason it is there is because of the properties that arise when you set it up that way.

From the series expansion, we can see that:

\(\displaystyle e^{tx} = 1 + tx + \frac{t^2x^2}{2!} + \frac{t^3x^3}{3!} + \cdots \implies \left[e^{tx}\right]_{t=0} = 1\)

\(\displaystyle \frac{d}{dt} \left( e^{tx} \right) = x + x^2t + \frac{x^3t^2}{2!} + \cdots \implies \left[ \frac{d}{dt} \left( e^{tx} \right) \right]_{t=0} = x\)

\(\displaystyle \frac{d^2}{dt^2} \left( e^{tx} \right) = x^2 + x^3t + \frac{x^4t^2}{2!} + \cdots \implies \left[ \frac{d^2}{dt^2} \left( e^{tx} \right) \right]_{t=0} = x^2\)

(etc.)

So the \(\displaystyle n \mbox{th}\) derivative of \(\displaystyle e^{tx}\) evaluated at \(\displaystyle t=0\) is \(\displaystyle x^n\)

This turns out to be convenient because the \(\displaystyle n \mbox{th}\) moment of a density function \(\displaystyle f(x)\) is

\(\displaystyle \int_{-\infty}^{\infty} x^n f(x) \, dx\)

So if we take

\(\displaystyle M_X(t) = E \left[e^{tX}\right]=\int_{-\infty}^{\infty} e^{tx} f(x) \, dx\)

we can basically isolate one of the moments by taking the proper number of derivatives and setting \(\displaystyle t=0\).