# Kinematics Question with weird units

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!

etotheipi
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

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berkeman
haruspex
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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.

etotheipi
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).

haruspex
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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."
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.
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?

etotheipi
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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