Dimensional Analysis exam

1. Sep 30, 2015

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

In a desperate attempt to come up with an equation to use during an examination, a student tries the following equations. Use dimensional analysis to determine which of these equations could be correct. Here x, v, and a, are the position, velocity and acceleration.

1. v2=ax
2. a=v2/x
3. x=v2/a
4. a=xv2
5. x=av2
6. v=ax
7. v=a/x
8. x=av
9. a=xv
10. v=a/t
11. v=at
12. v=t/a
13. a=v/t
14. t=av
15. t=v/a
16. t=a/v
17. a=t/v

2. Relevant equations

a=d/t^2
v=d/t

3. The attempt at a solution
This dimensional analysis thing is really killing me, I understand that the units have to match, thats about it I got out of my textbook and some youtube videos. How about to I apply that to solve this problem (T/F)

2. Sep 30, 2015

andrewkirk

Well, you've got two equations that express a and v as dimensions (ie in terms of d and t). All you need is the third that expresses position x as dimensions, which is of course x = d.

So now use those three equations to substitute everywhere for x, a and v into the list of 17 equations. The ones that balance (ie when both sides are the same when simplified) are possible. The others are not.

3. Sep 30, 2015

Staff: Mentor

Welcome to the PF.

The dimension of position is meters [m]. The dimension of velocity is meters per second [m/s]. The dimension of acceleration is meters per second squared [m/s^2].

The dimensions of the lefthand side (LHS) and the RHS must match. Multiply and divide dimensions the same way you multiply and divide numbers. If you end up with [m]/[m], those units cancel. If you end up with [m/s]/[m], you cancel the [m] units in the numerator and denominator, and are left with units of 1/[seconds], which is the same as Hz (Hertz, which is the unit of frequency, 1/[seconds]).

Does that help? Can you now show us the dimensions for the LHS and RHS for each of the questions in your post?

4. Sep 30, 2015

Staff: Mentor

The idea is to replace each variable with its fundamental units and see if the resulting equation balances. Fundamental units include:
[L] = Length
[M] = Mass
[T] = Time

So, for example, your first equation becomes:

$v^2 = a x \rightarrow ([L] [T]^{-1})^2 = [L][T]^{-2} [L]$

Do the algebra and see if it balances. Then see if you can do the rest.

5. Sep 30, 2015

Staff: Mentor

Piling on!

Now it's up to the OP...

6. Sep 30, 2015

So i just sub in units correct? Also two things, how can we assume d=x?
And the whole idea of why this works, what makes it so special if they are equal (rs=ls) in dimension value?

7. Sep 30, 2015

YEs thank you!

8. Oct 1, 2015

haruspex

To clarify, this L, M, T notation (and likewise Q for charge etc.) represents dimensions. Fundamental units are the base set of standard units laid down for the different dimensions according to some convention. Thus in modern SI units (MKS), the metre is the fundamental unit of length dimension, etc.

The square bracket notation does not appear to be completely standardised. A common usage is to put square brackets around a variable to signify the dimension of the variable (not around the dimensions themselves). Thus one may write for a velocity variable v: [v]=LT-1