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Average velocity, again

August 29, 2011

I’m thinking today about average velocity again, because on a certain day next week, I have to talk about average velocity. I have decided to talk a little about why average velocity isn’t typically found by adding up velocities and dividing that sum by the number of velocities you add up.

But on the very next day, I’m going to have to use this formula

Vavg = (vf + vi) /2

And I’m going to have to explain why in this case, it looks like we are just taking some velocities and dividing by the number of velocities we have–the very thing I said they shouldn’t do the day before.

So, it’s time to think about averages, again. Let’s say I have these numbers 1,2,3,4,5,6.

I can find the average of these numbers by taking (1+2+3+4+5+6) /6

But that is equal to 21/6 which is equal to 7/2 which is equal to (6+1)/2, which is just the first number added to the last number divide by two. This can, of course, be generalized to any sum of sequential numbers, and even equally spaced numbers.

This is a lot like Vavg = (vf +vi) /2, where you take the initial velocity add it to the final velocity and divide by two. This expression for the average velocity is true only for an object moving with a linearly changing velocity, similar to the way that 1,2,3,4,5,6 was a “linear” sequence of numbers.

I’m not sure it’s worth going into all this, but I just think it’s funny when you tell students not to do something, and the very next day you do that thing.

 

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3 Comments leave one →
  1. Andy "SuperFly" Rundquist permalink
    August 29, 2011 11:37 pm

    I think about it this way: v_\text{avg}=\frac{\text{distance}}{\text{time}}=\frac{\sum_i^N v_i dt}{\sum dt}=\frac{dt \sum v_i}{N dt}=\frac{\sum vi}{N} Doesn’t that amount to adding all the speeds up and dividing by the number? Of course, it assumes that all speeds are “on” for the same amount of time and so acceleration isn’t allowed.

  2. August 29, 2011 11:54 pm

    Nice Latex! Do you have to do anything special to use that with wordpress? I seem to remember someone having a blogpost about this.

    Yeah, that makes sense. Another way of thinking about it is this: One kind of “average” velocity is the time-weighted average. Given that trajectories must be continuous in euclidean space, the “time-weighted average” velocity is equal to the “equivalent velocity” described by displacement over time. There are kinds of spaces where time-weighted average velocity are not equal the “equivalent velocity” include the commonly used wrapping boundary conditions.

  3. August 30, 2011 12:13 am

    I’m not sure if your course incorporates calculus or not, but this could be an excellent opportunity to explore summing vs. integrating. Let’s say a car has a velocity of 1 at t=0 and 6 at t=5, and increases linearly. Integrating geometrically (a fancy way of finding the area of a triangle and rectangle, we get that the car has moved 24 units, rather than 21. (See visual aid here: twitpic.com/6dbgll .) We see the red step function that assumes v=1 along t=[0,1), and that one can place six blue triangles, area 1/2 each, to turn the step function into the triangle, accounting for the difference of 3. To relate this back to the post, the average of the numbers may be 21/6 (aka 3.5), but if we make a continuous function out of them, the average velocity is 24/6, more commonly known as 4.

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