What is the physical meaning of a Laplace Transform? Today I have learned what a Laplace Transform is and how it is defined. I have heard that it has wide applications in Physics/Engineering and so I started to wonder. I started with probably the simplest function in all of physics: distance formula in a straight line movement with no acceleration: 
$$ d(t) = vt $$
Where $d$ stands for distance, $v$ for velocity, and $t$ for time.
Taking Laplace Transform of this function leads to:
$$ \mathscr{L}\{d(t)\}(s) = \frac{v}{s^2} $$
Now as the $s$ shows up in the exponential in the definition of the Laplace Transform along with $t$ it seems to me that it has to have dimensions of $1/\text{time}$ for it to make any physical sense. This would suggest that $s$ is some sort of frequency, but I cannot understand what this new function tells us about the movement.
I understand that perhaps I'm using this transform in a wrong way/place and it has no real physical in that particular case. Therefore I would like to ask if there is any physical meaning in what I have done? If yes, what meaning? If no, why?
 A: My experience was that Laplace is typically taught as a toolkit, not explained. Which is a massive loss.
Laplace is a transform, allowing a function to be mapped to another like the usually more familiar Fourier transform. Whereas Fourier maps amplitude(time) into amplitude(frequency), using sin/cosine functions; Laplace is targeted at a function with a special property in calculus, $$f(x) = e^x$$
The special property is that "it is its own derivative", i.e.
$$ de^x/dx = e^x $$
That is the rate of change/gradient at any value x is the value f(x).
Just as in Fourier the transformed domain is the input of the transforming function (the frequencies of a sine function),


*

*the collection of all values of s define the set of
exponential functions that could be present as components in the
original function, 

*and the transformed function describes the actual
magnitudes with which they are present in the original function. 


The self-derivative property means that if you can transform a function using Laplace and the resultant function can usually be manipulated easily by calculus. Then it can be transformed back again to get the derivative or integral of the original function. It was a new toolkit for solving differential equations at the time.
It's rather like the much simpler log functions, if you take the $log_{10}$ of a number to multiply it by 10 you just need to add one; because the log function has special properties with respect to base 10. Only for logs the element undergoing "transformation" is a single number not a function.
A: Frequency Space and Fourier Transforms
The Laplace transform of a function $f(t)$ for $t>0$ is defined as,
$$F(s) = \mathcal L\{f(t)\}(s) = \int_0^\infty e^{-st}f(t) \, dt$$
for a complex variable $s$ with - as you correctly stated - the interpretation of a frequency. This in turn is related to the Fourier transform, if integrated over $(-\infty, \infty)$, for $s = i\omega$. However, note that generally substituting $F(i\omega)$ will not yield the Fourier transform $\hat f(\omega)$ as $i\omega$ is often a pole, reflecting the fact that the Fourier transform may have a delta function.
The Fourier transform of some signal $f(t)$ gives us a prescription on how to weigh different sinusoids of varying frequencies that make up the signal $f(t)$. Thus, the Laplace transform has the same interpretation, but instead we are interested in weights for representing it as a sum of exponentials.

Continuous Analogue of a Power Series
There is a mathematical motivation of the Laplace transform which nicely demystifies it further. To be specific, consider a function $f(x)$ with a power series representation,
$$A(x) = \sum_{n= 0 }^\infty a_n x^n.$$ 
We could ask the question: what is the relation between $a_n$ and $A(x)$? The answer is you perform the discrete sum. For example, $a_n = \frac{1}{n!}$ corresponds to $A(x) = e^x$.
We now ask, what is the continuous analogue of this prescription? Let us now consider instead of a discrete $a_n$, function $a(t)$ with $t\in [0,\infty)$ and instead of a sum we get an integral,
$$A(x) = \int_0^\infty a(t) x^t \, dt.$$
Thus to the continuous 'set of coefficients' $a(t)$, we can associate an $A(x)$. This integral has the best chance of converging if $x < 1$, as we are essentially taking higher powers as we sum. This motivates the substitution, $x := e^{-s}$ and we recover the Laplace transform,
$$A(s) = \int_0^\infty a(t)e^{-st} \, dt.$$
To convince yourself of this, take the Laplace transform,
$$\mathcal{L}\{\sin t\} = \frac{1}{1+s^2}.$$
Had we instead computed,
$$\int_0^\infty (\sin t )x^t \, \mathrm dt = \frac{1}{1+\log^2 x }$$
which is defined for $x < 1$, substituting $x= e^{-s}$ would recover the Laplace transform.
