Kinematics Question with weird units

[Ashelion]

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

 

[haruspex]

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]

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.