Kinematics Question with weird units

In summary, the A = 2.0 ms^(-5/2) would not affect my answers for parts (a) to (c), as I basically ignored the unit ms^(-5/2) and integrated the function to get the ones for velocity and displacement.
  • #1
Ashelion
1
0
Homework Statement
Hello I have a homework question that I am having trouble figuring out. It is in parts and I solved the first 3.

The acceleration of a particle is given by a = A√t where A = 2.0 ms^(-5/2).

(d) *Explain the dimensions of A.
Relevant Equations
I did parts (a) to (c) as per usual via integration and solved them for velocity, displacement, etc.
Would the A = 2.0 ms^(-5/2) affect my answers for parts (a) to (c)? Since I basically ignored the unit ms^(-5/2) and integrated the function to get the ones for velocity and displacement.

And as for part (d) explain the dimensions of A; I assumed that the ms^(1/2) comes from the square root due to the √t but I cannot simply do not know where the power -5 come from.

Thank you!
 
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  • #2
The equation is ##a = A\sqrt{t}##, now take the units of both sides, ##[a] = [A][\sqrt{t}]##, or explicitly $$\text{ms}^{-2} = [A] {\text{s}}^{\frac{1}{2}}$$Can you solve for ##[A]##?

*Caveat 1: As has been pointed out, the square bracket notation is not suitable for units.
*Caveat 2: As has also been pointed out, I have mis-interpreted the question
*Caveat 3: This means that this post is essentially useless, but I will keep it up for comedy value :wink:
 
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  • #3
etotheipi said:
The equation is ##a = A\sqrt{t}##, now take the units of both sides, ##[a] = [A][\sqrt{t}]##, or explicitly $$\text{ms}^{-2} = [A] {\text{s}}^{\frac{1}{2}}$$Can you solve for ##[A]##?
You are confusing units with dimensions.
##[a] = [A][\sqrt{t}]## leads to ##LT^{-2}=[A]T^{\frac 12}##.
This explains the dimension of A, as desired. It does not explain the units quoted, ##ms^{-\frac 52}##. Any other units of the same dimensionality could have been used.
 
  • #4
haruspex said:
You are confusing units with dimensions.
##[a] = [A][\sqrt{t}]## leads to ##LT^{-2}=[A]T^{\frac 12}##.
This explains the dimension of A, as desired. It does not explain the units quoted, ##ms^{-\frac 52}##. Any other units of the same dimensionality could have been used.

You are right, the notation ##[\dots]## is most commonly used for dimensions, but here I use it in the loose sense to denote 'the unit of' (the question says dimensions, but the book answer uses a unit so I had presumed they mixed it up themselves).

As you say, we have many possible choices for units (as opposed to dimensions) and we could have just as easily chosen $$\text{ft} \, (\mu \text{s})^{-2} = [A] \text{s}$$ which would also lead to an acceptable unit. Except of course the dimensionless coefficient ##\lambda_q## of our new unit in the full quantity ##q = \lambda_q u_X## will be different (but that is not so important to the question).
 
  • #5
etotheipi said:
the question says dimensions, but the book answer uses a unit so I had presumed they mixed it up themselves
No, the question is perfectly correct as given.
Consider a simpler example. "A car travels at a speed of 50mph. Explain the dimension given."
The dimension given is the dimension of the units expression 'mph', namely ##LT^{-1}##, which is appropriate for a speed. Had the question said "explain the units given" we would have had to explain not only the dimension but also why those particular units. A possible answer might include "because you (ok, we) Brits didn't have the sense to go metric fully."
etotheipi said:
we could have just as easily chosen $$\text{ft} \, (\mu \text{s})^{-2} = [A] \text{s}$$
I guess you meant $$\text{ft} \, (\mu \text{s})^{-2} = [A] \text{s}^{\frac 12}$$, but that makes no sense as an equation. The values of [A] are terms like L, T, LT-1 etc. It cannot equate to a combination of actual units.
etotheipi said:
Except of course the dimensionless coefficient ##\lambda_q## of our new unit in the full quantity ##q = \lambda_q u_X## will be different
Sorry, I don't understand what you are saying there. What dimensionless coefficient?
 
  • #6
OK I re-read the question and you are right that there is no mistake, and a simple ##LT^{-\frac{5}{2}}## will suffice. Next time I will read more carefully... I apologise for any confusion to the OP!

That said, I will try to explain what I was doing. Since the notation ##[\dots]## is conventionally used for dimensions, I will not use it here, because that will cause confusion (and furthermore it wouldn't be a function if we used it to denote 'the unit of', because there is not a unique result for each argument!).

A physical quantity ##p## is expressed as the product of a dimensionless coefficient ##\lambda_p## and a unit ##u_x## (we have infinitely many choices of unit for a given physical quantity, all of which share the same dimensionality). The quantity itself is independent of the unit i.e. ##p = \lambda_p u_x = \lambda_{p'} u_{x'}##.

Now expressing the original relation in this form, $$\lambda_a u_a = A \sqrt{\lambda_t u_t}$$ or rearranged $$A = \frac{\lambda_a}{\sqrt{\lambda_t}}u_a u_t^{-\frac{1}{2}}$$So in terms of the units in which we originally expressed ##a## and ##t##, our derived unit for ##A## is ##u_a u_t^{-\frac{1}{2}}##. Also since both ##\lambda_a## and ##\lambda_t## depended on our choice of units the dimensionless coefficient of the unit in the expression for ##A## (i.e. the number we calculate, without units) also depends on this choice of units.

I attach some notes in case I have explained anything badly - they use a different notation, ##\mathbf{A} = A\mathbf{a}## for the quantity, coefficient and unit respectively, but it amounts to the same thing.
 
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1. What are some examples of weird units used in kinematics?

Some examples of weird units used in kinematics include furlongs per fortnight, cubits per second, and leagues per hour. These units are rarely used in modern science and are mostly used for comedic or historical purposes.

2. How do you convert weird units to standard units in kinematics?

To convert weird units to standard units in kinematics, you can use conversion factors or online unit converters. Make sure to double check your conversions to ensure accuracy.

3. Why do some kinematics problems use weird units?

Some kinematics problems use weird units to challenge students and make them think outside the box. These problems can also be used to demonstrate the importance of unit conversions and how they can affect the final answer.

4. Are weird units ever used in real-life applications of kinematics?

In most cases, weird units are not used in real-life applications of kinematics. However, there may be some rare instances where these units are used in specialized fields or for specific calculations.

5. How can I practice solving kinematics problems with weird units?

You can find practice problems with weird units in kinematics textbooks or online resources. You can also create your own problems by converting standard units to weird units and vice versa.

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