Say I want to exchange some currency with someone and to agree on a fair rate we go to three different exchange places and average their rates. But if we were to get the rates the other way around we would end up with a different number. Which one would be fair?
Are you presuming the average is the arithmetic mean (add them all up and divide by how many you have)? Because there are many different averages out there and in this case you might consider the median.
Usually, “average speed” means total distance divided by total time.
If you travel 100 km/h for one hour and 150 km/h for one hour, you will have covered a greater distance during the hour you went 150 km/h than during the hour you went 100 km/h.
On the other hand, if you cover the same amount of distance at each speed, you will spend less time at the faster speed.
(If you are just given two rates of speed, without any other context, I don’t know whether there’s any way to meaningfully average them.)
Tom and Ray took separate road trips on the same day.
During the morning, Tom drove 100 km/h and Ray drove 75 km/h.
During the afternoon, Tom drove 150 km/h and Ray drove 125 km/h.
But Ray’s average speed for the whole day was greater than Tom’s (even though Tom drove faster than Ray in the morning and Tom drove faster than Ray in the afternoon).
Here’s how:
Tom drove 100 km/h for 4 hours, then 150 km/h for 1 hour, covering a total distance of 100x4 + 150x1 = 550 km in 5 hours, an average speed of 110 km/hr for the day.
Ray drove 75 km/h for 1 hour, then 125 km/h for 4 hours, covering a total of 75x1 + 125x4 = 575 km in 5 hours, an average speed of 115 km/h.
This is an example of Simpson’s paradox, which you can run into if you try to average ratios or rates.
Inuitively ? well whats wrong is that its generally meaningless to take the two instantaneous measurements and simply average them. Which two people (or person at different times) does things for the precise same length of time, or travels the precise same distance ? It does not take into account the distances actually travelled or times actually taken. What you really are going to do is consider how long far you can travel at 100km/h, and for how far you can travel at 150km/h , and then total up the distances and times, and then divide the two … you have your average for that trip. The full summation of distances and times is then invertable … whether you did it with a finite number of elements, or you did it with calculus (infinite numbers of elements ) you could integrate wrt distance or wrt time, or even mix up how you do different sections and get that same result speed as km/h , or its corresponding h/km.
inituitively, Your initial statement about averaging speed was only valid for just two events occurring for the same time period. If the events take different times( which neccessarily occurs for same DISTANCES), the average speed you gave is not a real thing.
Ok, I think I have found a way to understand it now.
Imagine two cars are trying to cover a distance as fast as possible. One car drives half the distance at 50 km/h and then the other half at 200 km/h, while the other car drives at 100 km/h for the whole distance.
But instead imagine that they are trying to cover as much distance as possible in a given time. One car drives at 50 km/h for and hour and then at 200 km/h for the next hour, while the other car drives at 100 km/h for two hours.