Dimensional Analysis of an object

In summary, the dimensions of A and B in the equation v=At^3 - Bt are [L/T^4] and [L/T^2] respectively. This is obtained by considering velocity as a measure of distance per time and working towards making the dimensions match through the use of proportions.
  • #1
future_vet
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Homework Statement


The speed, v, of an object is given by the equation v=At^3 - Bt, where t refers to time. What are the dimensions of A and B?

Homework Equations


None, but I know that to find dimensions, we look at the units that make up the variable, like the area of a room is [L^2]. I understand the basic concept, but have trouble applying it.

The Attempt at a Solution


v = At^3 - Bt.
I would think that since velocity is a measure of distance per time, we should get t to divide the distance. Could it be then that A is [L/T^4] and B= [L/T^2]?
I can only figure out this way to get T as the denominator in a way that could get A and B added up to get [L/T] for the answer...

Thanks for your help!
 
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  • #2
Yes, this looks right because you know velocity should have dimensions of displacement/unit time. So you need a displacement in each of the coefficients, and also units of time that gives an inverse proportion in the coefficients.

So, even though this can be done mentally, for other problems you would start with your original proportion and work towards making into the dimension you want; like so:

[tex] A*t^3 = \frac{L}{t}[/tex]

Now divide both sides by t^3.

[tex] A = \frac{L}{t^4} [/tex]

Or if you wanted to convert from km/h to m/s, you know that 1000m=1km and 60s=1m and 60m=1h, so:

[tex]\frac{km}{h} * \frac{1h}{60m} * \frac{1m}{60s} = \frac{1}{60^2} \frac{km}{s} [/tex]

So you see that the hours cancel and make minutes, then the minutes cancel and make seconds in the denominator. Now take care of the km units:

[tex]\frac{km}{h} * \frac{1h}{60m} * \frac{1m}{60s} * \frac{1000m}{1km} = \frac{1000}{60^2} \frac{m}{s} [/tex]

Just write out your proportions and see that the units cancel to get what you want.
 
Last edited:
  • #3
Thanks a lot!
 

1. What is dimensional analysis?

Dimensional analysis is a mathematical technique used to analyze and relate physical quantities. It involves converting units of measurement and ensuring that the units on both sides of an equation are equivalent.

2. Why is dimensional analysis important?

Dimensional analysis is important because it allows scientists to check the validity of equations and calculations. It also helps in simplifying complex equations and reducing errors in experimental data.

3. How is dimensional analysis used in science?

Dimensional analysis is used in science to convert units of measurement, such as length, mass, and time, into different systems of measurement. It is also used to check the consistency of equations and to derive new relationships between physical quantities.

4. Can dimensional analysis be used in all branches of science?

Yes, dimensional analysis can be used in all branches of science, including physics, chemistry, biology, and engineering. It is a fundamental concept that is applicable to any physical quantity that can be measured.

5. Are there any limitations or drawbacks to dimensional analysis?

One limitation of dimensional analysis is that it cannot be used to derive dimensionless constants, such as the gravitational constant or the fine structure constant. It also assumes that all physical quantities can be measured using a single set of base units, which may not always be the case.

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