We all know the average velocity (average speed, really) of an object is the total distance divided by the total time. If, for example, an object moved 20 meters in 4 seconds, it had an average speed of 5 meters/second during that time. This is true even if the speed was not constant.
But what’s the proof for this? Is there a calculus theorem that says the average speed of an object is the total distance divided by the total time? The mean value theorem? Or is it simply a definition?
It’s just the definition of ‘average’, i.e. the single constant velocity that covers the given distance in the given time.
ETA: I suppose, if you had the displacement as a function of velocity and time, i.e. something like d(t) = v(t)*t, you could derive some kind of expression for the mean velocity involving just the function of velocity, but it works out to the same thing in the end.
It is the definition of average rate, which applies to speed as well as any other this/that rate (gallons/minute, joules/sec, etc.).
A very similar question could be asked, and answered, about the exact meaning (or definition) of “area”. We all know the grade-school formulas for areas of several simple shapes (circle, square, rectangle, triangle…). They just give you a number, and we learn to just accept those numbers as what area is.
Now try that for irregular regions with curvy boundaries. Here, we start with only an “intuitive” idea of what area must mean (having something to do with how much paint you would need to paint it). Then we develop integration formulas that seem to give answers consistent with our intuition.
It wasn’t until I got into Differential Equations (and towards the end of the semester too) that we learned the Cold Hard Truth. We learned double-integration techniques for various kinds of bounded regions. (Double integration means you don’t just chop up the region into narrow slices, but you do so both horizontally and vertically, chopping up the region into little squares.) Only then did we learn that the measure we call Area is, by definition, nothing more nor less than whatever you get by doing that.
Then we used double integration techniques to develop formulas for areas of squares, circles, triangles, etc. Well, surprise, surprise! The formulas one derives turn out to be none other than those formulas we all learned in fifth grade!
Instantaneous velocity (v) is the change in position divided by some infintesimal piece of time (dt).
If I wanted to take an average of all these instantaneous velocities over some period of time T, I’d multiply each instantaneous velocity by the amount of time it was traveling at that velocity (dt), and than sum the results, and divide by T. But v*dt is just the distance travelled in dt, so the sum over the whole time period is the total distance travelled (D). So our average velocity is just D/T.
Simplicio has it. It is essentially the content of the Second Fundamental Theorem of Calculus.
Note that total distance over total time is the average of velocities weighted by the time spent at each velocity. If one took a different kind of weighting in their averaging of velocities (e.g., weighting each velocity by the distance spent upon it, as might be relevant for, say, understanding the wear effected on a road), the result is no longer guaranteed (or even likely) to be total distance over total time.
I’m 30 never went to school but have a few questions please don’t judge if they sound displaced.first is about the use of arc light and laser pulses(together).second is about the post that brought me to this message board something about percieved light spectrum if the authors of that post are still around please reference to this post I guess these will be all (TOE) questions with only my own mechanical logic to back it.:smack:
By that definition, a body moving in circle will have zero speed - if you consider one revolution.
Nit Pick - Speed and velocity are used interchangeably in this post but it is not so. The latter is a vector
Yea, that’s the answer I was looking for. Thanks Indistinguishable and Simplicio.
I should know this stuff, seeing how I took Calc I-IV. But that was centuries ago. 
You didn’t nitpick enough. Just as speed is a scalar but velocity is a vector, so too distance is a scalar, but displacement is a vector. If you travel once around a circle, your displacement is zero, but your distance is the circumference, and your speed is in fact your distance divided by your time.
As I started to read the OP I thought it was going to be about the Mean Value Theorem. I.e., as long as there is no “jump” in speed, at at least one point the instantaneous speed is the same as the average speed. That actually requires some Real Math.
This example also shows that although speed is considered to be the magnitude of the velocity vector, ‘average speed’ is not the magnitude of the average velocity. There may well be circumstances where you’d want to use that variation, but it’s almost always more reasonable to define speed, or at least average speed, in terms of distance and time.