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.

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.