Average Speed and Average Power confusion

Click For Summary

Discussion Overview

The discussion revolves around the concepts of average speed and average power, focusing on their mathematical definitions and interpretations. Participants explore how these averages are calculated and the implications of using integrals in their definitions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant expresses confusion about the mathematical understanding of average speed and average power, questioning how the general idea of averaging applies to these concepts.
  • Another participant suggests that understanding integrals in discrete terms may help clarify the concepts of average speed and average power.
  • A third participant presents a mathematical formulation for the time-weighted average of a quantity Q(t), illustrating how average speed can be derived from integrals over time.
  • The same participant provides examples of how average velocity and average speed can be expressed using integrals, emphasizing the role of time in these calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the understanding of average speed and average power, and multiple viewpoints regarding the mathematical treatment of these concepts remain present.

Contextual Notes

The discussion includes references to integrals and their application in defining averages, but does not resolve the underlying assumptions or the clarity of these mathematical concepts.

LLT71
Messages
73
Reaction score
5
I found it somehow tricky to understand average speed and average power in mathematical fashion.
I suppose general idea of taking average of some quantity is: sum of all values/number of values.
how that idea fit's into idea of average speed=distance traveled/time traveled

similarly, average signal power:
signal_power.PNG

why is it over T?
 
Physics news on Phys.org
LLT71 said:
I found it somehow tricky to understand average speed and average power in mathematical fashion.
I suppose general idea of taking average of some quantity is: sum of all values/number of values.
how that idea fit's into idea of average speed=distance traveled/time traveled

similarly, average signal power:
signal_power.PNG

why is it over T?

Do you understand integrals in discrete terms?

https://betterexplained.com/articles/a-calculus-analogy-integrals-as-multiplication/
 
  • Like
Likes   Reactions: LLT71
time-weighted average of Q(t)=##\displaystyle\frac{\int_0^T Q(t)\ dt}{ \int_0^T \ dt }=\frac{\int_0^T Q(t)\ dt}{ T } ##
(For example, the average-velocity is a time-weighted average of velocity:
##\displaystyle\frac{\int_{t_1}^{t_2} v(t) \ dt}{ \int_{t_1}^{t_2} \ dt }=\frac{\int_{t_1}^{t_2} \frac{dx(t)}{dt} \ dt}{ \int_{t_1}^{t_2} \ dt }=\frac{x(t_2)-x(t_1)}{ t_2-t_1}=\frac{\Delta x}{ \Delta t} ##.

the average-speed is a time-weighted average of speed:
##\displaystyle\frac{\int_{t_1}^{t_2} |v(t)| \ dt}{ \int_{t_1}^{t_2} \ dt }=\frac{\int_{t_1}^{t_2} |\frac{dx(t)}{dt}| \ dt}{ \int_{t_1}^{t_2} \ dt }=\frac{d}{ t_2-t_1}=\frac{d}{ \Delta t} ##.
)z-weighted average of Q(z)=##\displaystyle\frac{\int_0^Z Q(z)\ dz}{ \int_0^Z \ dz }=\frac{\int_0^Z Q(z)\ dz}{ Z }##
(For example, the center of mass is a mass-weighted average of position:
##\displaystyle\frac{m_1x_1+m_2x_2+m_3x_3 }{m_1+m_2+m_3}=\frac{m_1x_1+m_2x_2+m_3x_3 }{M}##.
If all of the "masses" are equal (so each item has equal "weight" in the average),
then the mass-weighted average reduces to
the count-weighted (ordinary, "straight") average:
##\displaystyle\frac{(1)x_1+(1)x_2+(1)x_3 }{(1)+(1)+(1)}=\frac{x_1+x_2+x_3 }{3}##.
)
 
  • Like
Likes   Reactions: LLT71
thanks guys!
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
6K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K