Powers of p and q from a=v^p*t^q

  • Thread starter Thread starter valerieangel
  • Start date Start date
AI Thread Summary
To determine the powers of p and q in the equation a = v^p * t^q for dimensional consistency, it is essential to analyze the units involved. Acceleration (a) is measured in m/s², while velocity (v) is in m/s and time (t) in seconds. The only values that satisfy the dimensional equation are p = 1 and q = -1. This ensures that the dimensions on both sides of the equation match correctly. The discussion emphasizes the importance of unit analysis in solving for the powers in physical equations.
valerieangel
Messages
10
Reaction score
0
How do I go about solving for the powers of p and q to make the equation dimensionally consistent?

Any help or advice would be much appreciated! :smile:
 
Physics news on Phys.org
I'm assuming v is velocity, t is time and a is acceleration?

Acceleration has units of m/s2, while velocity has units of m/s. So the only valid values of p and q would have to be p = 1 and q = -1.
 
dipole said:
I'm assuming v is velocity, t is time and a is acceleration?

Acceleration has units of m/s2, while velocity has units of m/s. So the only valid values of p and q would have to be p = 1 and q = -1.

Yes, I failed to mention that. Thank you for your help. :smile:
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
View full post »

Similar threads

Solving system of 3 quadratic equations
Replies
4
Views
2K
If p is even and q is odd, then ##\binom{p}{q}## is even
Replies
19
Views
3K
Q unless negate p = if p then q
Replies
5
Views
2K
Calculating Diagonals of a Polygon: What are the Values of p and q?
Replies
17
Views
3K
What Is the Correct Symbolic Form and Truth Table for ~(p|q)?
Replies
6
Views
2K
Direct Proof that every zero of p(T) is an eigenvalue of T
Replies
24
Views
3K
I Adding total time derivative to Lagrangian/Canonical Transformations
Replies
2
Views
218
How Can I Solve This Equation to Find the Correct Expression?
Replies
7
Views
2K
The infinite limits of the probability transition matrix for Markov chain
Replies
24
Views
4K
A Deriving the perturbative expansion from Hubbard to Heisenberg
Replies
0
Views
2K

Hot Threads

Recent Insights

Back
Top