Find powers p and q to make A=V^pT^q dimensionally consistent

[swatmedic05]

Acceleration is related to velocity and time by the following expression: A=V^pT^q. Find the powers p and q that make this equation dimensionally consistent. I already got the answer.I guessed and got lucky
Could someone tell me how to get the answer step by step
Please and Thank You

 

[rock.freak667]

According to your https://www.physicsforums.com/showthread.php?t=424587", you got it out.

Any reason why you are having trouble again?

 

[swatmedic05]

I thought I understood it but I guess i didnt

 

[rock.freak667]

I thought I understood it but I guess i didnt

Well post your work up to where you got confused.

 

[swatmedic05]

The exponents from when u posten last time was 2 and -1 where did those come from

 

[rock.freak667]

The exponents from when u posten last time was 2 and -1 where did those come from

Ah, remember your rules of indices?

A= L/T² = LT^-2

V=L/T=LT^-1

t=T

So you'd have

LT^-2= (LT^-1)^p(T)^q

Remember these rules:

(ab)ⁿ=aⁿbⁿ and (aⁿ)^m=a^mn and that (aⁿ)(a^m)=a^m+n

It is just applying those rules.

 

[swatmedic05]

whats indices I don't remember that?

 

[rock.freak667]

whats indices I don't remember that?

exponents, powers...you know things like (t)(t) =(t¹)(t¹)=t², the 1s are called 'indices'.

http://www.maths.abdn.ac.uk/~igc/tch/engbook/node5.html"

 

[swatmedic05]

so will the answer alway be 1 and -2

 

[rock.freak667]

so will the answer alway be 1 and -2

No you need to simplify the terms on the right side of the equation until you just have one L and one T on the right.

 

[swatmedic05]

I get it now you get the negative by dividing the numbers and subtracting the exponents

 

[rock.freak667]

I get it now you get the negative by dividing the numbers and subtracting the exponents

Well not by subtracting, but by the rule that 1/aⁿ=a^-n

 

[swatmedic05]

Ok I think I get it now, now i hope he gives us a practice problem before the test

Thank you again

 

[rock.freak667]

Ok I think I get it now, now i hope he gives us a practice problem before the test

Thank you again

You could try this problem and say what you got for 'p' and 'q'.

 

[swatmedic05]

what problem the one I just gave as an example

 

[rock.freak667]

what problem the one I just gave as an example

Try S=A^pt^q

Where S= displacement/distance , A = acceleration and t= time.

 

[swatmedic05]

So is S= [L]

 

[rock.freak667]

So is S= [L]

Yes.

 

[swatmedic05]

So the equation is:
[L]=[L]/[T]^2*[T]^2

 

[rock.freak667]

So the equation is:
[L]=[L]/[T]^2*[T]^2

Right, now simplify the right side.

EDIT: you forgot the powers 'p' and 'q'

 

[swatmedic05]

would the right side be four(4)

 

[rock.freak667]

would the right side be four(4)

uhm no, you need to simplify the right side

(L/T²)^p(T²)^q such that you will have just one L and one T i.e. group all the Ls and all the Ts together.

 

[swatmedic05]

im sorry to be taking up all ur time

so u would combine like terms [T]

 

[rock.freak667]

im sorry to be taking up all ur time

so u would combine like terms [T]

Yes you would. So the L on the right would just become L^p, what does the T become?

 

[swatmedic05]

T^q i think

 

[rock.freak667]

T^q i think

Remember you have

(L/T²)^p(T²)^q

Clearly (T²)^q becomes T^2q right? What does (L/T²)^p become? [remember (a/b)ⁿ=aⁿ/bⁿ]

 

[swatmedic05]

(L/T^2)^p becomes
(L^p/T^2p)

 

[rock.freak667]

(L/T^2)^p becomes
(L^p/T^2p)

So then the right side becomes (L^p/T^2p)(T^2q) and grouping the Ts together, what does the entire right side become?

 

[swatmedic05]

would it be -2 2

 

[rock.freak667]

would it be -2 2

Are there any 'p's and 'q's ?