Dimensional analysis of this equation

In summary: So, in summary, the equation x-x0 = v0(t) + (1/2)a(t^2) contains terms with units of meters, seconds, and meters per second. Therefore, the units of v0 must also be meters per second in order for the equation to be dimensionally correct.
  • #1
uchicago2012
75
0

Homework Statement


In the equation below, x and x0 are distances in meters, t is the time in seconds, and v0 is a speed in meters per second. Find the units of a.
x-x0 = v0(t) + (1/2)a(t^2)
Find the dimensions (or units) of v0 that will make this equation dimensionally correct.


The Attempt at a Solution


So I found that a is acceleration, or m/s^2, but the last question confuses me- possibly I don't understand the wording? When I went through the equation and subbed in the units, I got:

m-m = (m/s)(s) + (1/2)(m/s^2)(s^2) which comes down to
m= m + (1/2)m = (some number)m

So what exactly is the problem with the units or dimensions or whatnot? Or is the point for us to realize there isn't one? Or perhaps I found a's units wrong.

Thanks in advance for any help.
 
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  • #2


The last question is "Find the dimensions (or units) of v0 that will make this equation dimensionally correct." They just want you to find the units of v0.

Why would you think you might have found the units of a wrong? What else could they possibly be?
 
  • #3


But didn't they give us the units of v0? As in meters per second?

Sorry if this is a stupid question, but I don't understand why they're asking for what I think they gave us.
 
  • #4


Oh yeah, they do... I'm not sure why they do that either. Maybe whoever wrote the question just wasn't paying attention.
 
  • #5


uchicago2012 said:
But didn't they give us the units of v0? As in meters per second?
Yes, they gave you units for v0, but are they the right units?
 
  • #6


Uh, I think so? Otherwise the first term wouldn't be in meters, which would be a problem. Wouldn't it? Ugh.
 
  • #7


Yes, that would be a problem, if the first term (or any term) didn't have the same units as all the other terms.
 

1. What is dimensional analysis?

Dimensional analysis is a mathematical approach used to analyze the relationships between different physical quantities in a given equation. It involves identifying the dimensions and units of each quantity and using them to determine the overall dimensions and units of the equation.

2. Why is dimensional analysis important?

Dimensional analysis is important because it allows scientists to check the validity of equations and ensure that they are physically meaningful. It also helps in converting between different units and simplifying complex equations.

3. How do you perform dimensional analysis?

To perform dimensional analysis, you must first identify the dimensions and units of each quantity in the equation. Then, you can use dimensional analysis rules, such as the equality of dimensions and the principle of dimensional homogeneity, to determine the overall dimensions and units of the equation.

4. What are some common applications of dimensional analysis?

Dimensional analysis is commonly used in physics, chemistry, engineering, and other scientific fields to check equations, convert between units, and solve problems involving physical quantities. It is also used in scaling laws, which describe the relationship between different physical systems.

5. Can dimensional analysis be used for all types of equations?

No, dimensional analysis can only be used for equations that involve physical quantities with measurable dimensions and units. It cannot be applied to purely mathematical or abstract equations that do not have physical meaning.

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