What are the dimensionally consistent powers in the expression v = a^p * t^q?

[tony873004]

Here's a homework question I have. I don't even understand what they're asking. Does somebody know the answer, and can you explain it in a way that I can do the other problems like it?? Thanks in advance for your help

Velocity is related to acceleration and time by the following expression, v = a^p * t^q. Find the powers p and q that make this equation dimensionally consistent.

 

[enigma]

Hi tony,

In the future, questions like this should be posted in the homework help forums.

What the question is asking you is to try to make the units multiply out correctly. It is one of the easiest ways to ensure that you've done a problem correctly.

Instead of using numbers to multiply together, you want to multiply the units together.

You know that velocity has units of length/time. You know that acceleration has units of length/time^2. If you treat the units as variables, you should notice what values of p and q which make the equality sign true.

Give it a shot... write back if you're still stuck.

 

[tony873004]

Hi tony,

In the future, questions like this should be posted in the homework help forums.

I just figured that out! Sorry, it was my first post

The left side of the = is L / T^2. Does this mean that I have to make the right side equal L / T^2 ? If that's the case, then (L / T) should have a power of one, and T should have whatever power allows me to combine it with the T in the L/T denominator to get T^2.

L / T^2 = (L / T) ^ p * T ^q

I asked the teacher this question after class today, he looked at it quickly and said that p =1 and q = -1.

But we do our homework on Webassign.net, which instantly tells you if your answers are correct. The p=1 was correct, but the q=-1 wasn't. Last time I ask my teacher for help

I guess I just forget the algebra that let's me move the T from outside the parenthesis to the denominator of the expression inside the parenthesis.

(p.s. I saw on someone else's post that there's a nicer way to say stuff like T^q. A way that makes it look more textbook by actually placing the q as a superscript of T. How did he do that?)

 

[Grizzlycomet]

(p.s. I saw on someone else's post that there's a nicer way to say stuff like T^q. A way that makes it look more textbook by actually placing the q as a superscript of T. How did he do that?)

$$T^q$$
Its called Latex. Check out this page to see how to make it
https://www.physicsforums.com/showthread.php?t=8997

Oh, and welcome to PF!

 

[Mk]

Oh, and welcome to PF!

Yes! Welcome!

 

[JasonRox]

Sometimes even if you forget the equations, you can get the answer simply by looking at the dimensions, and working it out. Of course it has to make sense, but it is a good step towards understanding.

 

[Galileo]

I asked the teacher this question after class today, he looked at it quickly and said that p =1 and q = -1.

Your teacher screwed up.

$$v=a^pt^q$$
Velocity $v$ is measured in meters per second (m/s).
Acceleration $a$ is measured in meters per second squared $(m/s^2)$.
Time $t$ is ofcourse measured in seconds (s).
So the left side is m/s.

 

[arildno]

Let's look at how you can directly compute p and q using "unit maths"
$$[v]=L^{1}T^{-1}$$
$$[a]=L^{1}T^{-2}$$
$$[t]=T^{1}$$
Hence, we require:
$$[v]=[a^{p}t^{q}]=[a^{p}][t^{q}]=[a]^{p}[t]^{q}$$
Or, filling in:
$$L^{1}T^{-1}=L^{1*p}T^{-2*p}T^{1*q}=L^{p}T^{q-2p}$$
Now, L and T are independent dimensions, so the powers must equal each other separately:
L:
1=p
T:
-1=q-2p

Hence, you may solve for: p=1, and q=1

 

[HallsofIvy]

Your teacher screwed up.

$$v=a^pt^q$$
Velocity $v$ is measured in meters per second (m/s).
Acceleration $a$ is measured in meters per second squared $(m/s^2)$.
Time $t$ is ofcourse measured in seconds (s).
So the left side is m/s.

It does not follow that the teacher "screwed up". In his post, tony873004 asserted that "The left side of the = is L / T^2" which was incorrect. If he gave that to his teacher when he was asking the problem, then his teacher was correct. I suspect he didn't show his teacher the original problem.

 

[HallsofIvy]

tony873004: when you are asking us or your teacher for help (and you should certainly continue doing both) be sure to state the ENTIRE PROBLEM and show what you have already done.

 

[tony873004]

Arildno's explanation made sense to me. It's still going to take a little time to digest this, but p=1 and q=1 are correct according to Webassign.net.

I did post the entire question. The $$L / T^2$$ was my screw-up, not the teacher's, but I made this mistake after talking to the teacher. He did see the assignment straight from the book.

Thanks everyone for helping. This is so awesome that there's a forum of people willing to help! Now I've got a $$26/26$$ on this assignment .

And TEX is very cool