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The concept of a "rate"

 
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Mar19-13, 02:07 AM   #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.
 
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Mar19-13, 05:32 AM   #2
 
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Hi Square1!
Quote by Square1 View Post
… I can get the 'rate' of the curve by using the definition of slope like we were taught Δmeters/Δtime ie the speed.

… I just don't understand what division has to do with it - multiple subtractions of units of time, from the numerator??
The slope is a division: units up over units across.

(and i don't understand the last part of your question, about subtractions)
 
Mar19-13, 07:55 AM   #3
 
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"per" means "divide".
 
Mar19-13, 01:37 PM   #4
 
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The concept of a "rate"


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.
 
Mar19-13, 04:15 PM   #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.
 
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