Dimensional correctness of equations I have studied that a dimensionally correct formula/equation may or may not be correct. But in order for a formula or an equation to be correct, it must be dimensionally correct, according to the principle of homogeneity. 
What about so many $x$-$t$, $v$-$t$, $a$-$t$, and $a$-$v$ relations then? 
For example, displacement as a function of time is given by 
x = 2t, or 
x = t² - t ?? 
Likewise, v = 3t, 
v = t², 
a = 1/x² 
...... etc. Just a few examples. 
In all of the above examples, from what I see, they are dimensionally incorrect. It's pretty obvious to figure it out, isn't it? My question is, why are they being used in my books then? They show up often in illustrations, or when I solve numerical problems. 
EDIT : I see my question has been marked as a duplicate. I'm totally surprised. Because my question is different, from the one asked in the other thread. I know that the equation for working out the displacement in nth second is dimensionally correct, there's a hidden (1s) in three places on the right hand side. And no, we're not choosing this (1s) arbitrarily. It can be proved using the 2nd equation of motion, that there's a hidden (1s) in three places on the right hand side. I already know, that equation is dimensionally correct. What I've asked is completely different. I mean, the answers I got, were something like, there's a hidden 1 m/s², or 1 m/s. How do I know for sure that the hidden constant has the same dimensions as m/s² or m/s, or something else? Is there a proof? Because as I said, the equation for working out the displacement in nth second is dimensionally correct and can be proved using the 2nd equation of motion, that there's a hidden (1s) in three places on the right hand side. We're not just choosing this (1s) ''arbitrarily'' just to make the equation dimensionally correct.
 A: All of those equations seem to have incorrect dimension because the dimensions of the constants in them are not explicitly specified. For example, for $x(t)=2*t$, by taking the derivative you can check that $\frac d{dt}x(t)=v(t)=2$. So the 2 in the equation actually has dimension of velocity, and then you have that the dimension of x is:$$dim(x)=\frac {distance}{time}*time=distance$$
So the equation has the right dimensionality, the apparent problem comes from omitting the dimensions of the constant "2". And the same argument can be used for all the equations that you presented.
ADDED:
I think you need to go back and try to understand what you mean by the equation $x(t)=2t$ or $x(t)=t^2$ in the first place. You can look at it as a simple function of one argument, with no physical meaning, and then that 2 and the 1 are just numbers. Or you can be trying to model some physical situation with the equation. If what you are trying to model with $x(t)=2t$ is the position of a particle as a function of time, then x has to have dimension of length, just as t must have dimension of time, and because of that 2 has to have dimension of velocity, because if not, the equation would be meaningless. The 2 in the equation does not show dimensions explicitly because it would just clutter notation, and it would be useless to cary it around when doing calculations.
Dimensional analysis is important, but is trivial in these simple self-explanatory cases, no information can be taken from it. The real importance of dimensional analysis comes when using fundamental constants, like $c$ or $h$, because then there might be unic, or at least a few, ways of combining these to get the units of some quantity you are trying to find. 
