# Unit Analysis

1. Jan 24, 2010

### mistalopez

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

The general equation for a parabola is y=ax²+bx+c, where a, b, c are constants. What are the units of each constant?

2. Relevant equations

y=ax²+bx+c

3. The attempt at a solution

The answer is a: 1/m; b: dimentionless; c: m

How exactly did they get that answer in the book? Can someone explain? I do not understand how b is dimensionless rather than a which is a fraction. Also, where did the 1/m come from?

2. Jan 24, 2010

### rl.bhat

Hi mistalopez, welcome to PF.
y is the displacement which is measured in meter.
You must have the same unit in right hand side.
So ax^2 = 1/m*m^2
Similarly the other two.

3. Jan 24, 2010

### mistalopez

Thanks for such a quick reply and the warm welcome! However, the answer flew way above my head. Would it be possible to explain it in more depth for someone who does not have much experience in physics to understand?

4. Jan 24, 2010

### Char. Limit

The units on both sides of the equation must match. So since y is in meters, (and I believe x is in meters) $$ax^2$$ must be in meters, and so on for the other two.

So, just for ax^2, you have meters squared times some unit gives meters. So, the unit for a must be inverse meters. I'll let you handle the other two.

5. Jan 24, 2010

### ideasrule

First, the constants in a parabola equation have no units. A parabola is a mathematical construct, not a physical one, so it doesn't have units any more than the number 2.1 does.

Second, I suspect the parabola is supposed to represent a free-fall trajectory, where y represents height and x represents horizontal displacement. In that case, y must have units of meters, so ax^2, bx, and c must all have units of meters. Otherwise, you couldn't add them; what does it mean to add 2 m to 3 s, for example?

Since x must have units of meters, what must the units of "a" be to make ax^2 also have units of meters? How about bx?