1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The concept of a rate

  1. Mar 19, 2013 #1
    The concept of a "rate"

    Here's another question from good ol square :|

    I was hoping to get some clarification about the concept of a rate of two quantities, with different units, like speed [itex]\frac{a-units}{b-units}[/itex]. How does the expression for division [itex]\frac{a-units}{b-units}[/itex] arise from the statement like 'a' meters per 'b' seconds?

    If I were to plot meters vs. time for some object moving at constant speed, I can get the 'rate' of the curve by using the definition of slope like we were taught Δmeters/Δtime ie the speed.

    I also understand that acceleration is the change in velocity, 'per' given desired time interval ie Δvelocity/Δtime.

    In both cases, I don't think I have conceptual issue with why it makes sense to define these rates of change with respect to time, I just don't understand what division has to do with it - multiple subtractions of units of time, from the numerator?? :S Thanks.
     
  2. jcsd
  3. Mar 19, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Square1! :smile:
    The slope is a division: units up over units across.

    (and i don't understand the last part of your question, about subtractions)
     
  4. Mar 19, 2013 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    "per" means "divide".
     
  5. Mar 19, 2013 #4
    A ball rolls 5 feet every 30 seconds. How far has it rolled after 3 minutes (solve without using fractions)? How did you reason out your answer? Is it a bit like "There are 5 feet attributed to every slice of 30 seconds, so I just have to find how many slices of 30 seconds fit into 3 minutes" ? That's pretty much the motivation for a fraction: one property is directly associated to a certain amount of a different property, so operations with that property correspond to operations on the proper corresponding multiple of the first property.
     
  6. Mar 19, 2013 #5
    Thank you for the feedback everyone. I am backing out of this question though. I don't know if I am overthinking it or something, but I am having a "I'm not exactly sure what my own problem even is" moment. Although, there is something I'm not getting :S Maybe I'll come back to it another time.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: The concept of a rate
  1. Concept of linearity? (Replies: 3)

  2. Interesting Concepts (Replies: 1)

  3. Infinity a concept? (Replies: 4)

  4. Concept of volume (Replies: 6)

Loading...