# Dimensional analysis - working out if this is dimensionally correct

1. Jul 26, 2012

### ulfy01

1. The problem statement, all variables and given/known data

I'm trying to work out if the following is dimensionally correct. I think I'm getting stuck at the (x - x0)

2. Relevant equations

In this case v is velocity (L/T), a is acceleration (L/T²), and x represents displacement, which is a length (L)

3. The attempt at a solution

My attempt was such:

v² = v0² = (L/T)² or L²/T²

a = L/T²

(x - x0) confuses me. That would work out to (L - L), correct? So a(x - x0) = L/T²(L - L)

Don't (L - L) cancel out, leaving me just with L/T² which is NOT the same as L²/T²?

According to the sheet, this expression is supposed to be dimensionally correct. Any pointer appreciated.

2. Jul 26, 2012

### ehild

Hi ulfy1, welcome to PF!

The difference between two lengths is length. Just think, what you get when you cut a 2 meter long piece from a 10 m long string. You get a piece of 8 meter length.

ehild

3. Jul 26, 2012

### ulfy01

Ah, I think I got it! I have to get that dimensions are just that, dimensions.

So in truth, a(x - x0) is really just L/T²(L) which is L²/T²

I think that's the right conclusion and makes the expression correct.

Thanks!