Here is a textbook situation in Time, Speed and Distance: A man goes from point A to point B with velocity v1 and returns with velocity v2. What is his average velocity?
Using the formula , the average velocity can be found to be . So far so good.
Ever wondered why we get this result? Why do we get the velocity as the harmonic mean of the two velocities? The answer lies in basics of arithmetic and harmonic progressions.
Let t1 and t2 be the time taken while going from A to B and coming back. The situation is shown below:
Now what will a hypothetical average velocity in this situation mean? It will mean that a person takes the same time, taverage, while going from A to B and coming back. The situation is summarized below:
The total time taken will be same as t1 + t2
--> 2 × taverage = t1 + t2 or taverage = (t1 + t2)/2
In another words, t1, taverage, and t2 will be in arithmetic progression.
So how is this related to velocity?
Remember that when distance is constant velocity is inversely proportional to time?
I.e. V is proportional to 1/T or V = k/T.
If T1, T2, and T3 were in arithmetic progression, then 1/T1, 1/T2, 1/T3 are in harmonic progression => k/T1, k/T2, k/T3 are in harmonic progression => V1, V2 and V3 are in harmonic progression!
It can also be proved that if V1, V2 and V3 are in arithmetic progression, then T1, T2, and T3 are in harmonic progression.
So here’s the rule:-
Now let’s apply these rules in practical problems:-
Re: Quant Exotica- Use of Arithmetic and Harmonic Progressions in Time, Speed, and Distance