Updating average speed with latest observation of distance and time

Question

Suppose I've traveled a cumulative distance $D$= 39.190 meters in time $T$ = 2.00 secs.

My average speed is then simply:

$$ S = \frac{D}{T} = \frac{39.190\ m}{2.00\ sec} = 19.595\ \frac{m}{sec}. $$

Suppose in the next observation interval, I go $d$ = 0.788 meters in $t$ = 0.02 seconds. My new average speed is then:

$$ S' = \frac{D'}{T'} = \frac{(D + d)\ m}{(T + t)\ sec} = \frac{(39.190 + 0.788)\ m}{(2.00+0.02)\ sec} = 19.791\ \frac{m}{sec} $$

I'm trying to find a way to calculate $S'$ from $S, d, \textrm{and}\ t$. That is, I want to derive a function $f$ such that $S' = f(S, d, t)$.

If I assume that the speed $S$ is normalized to be $S = \frac{S\ m}{1\ sec}$, then

$$ S' = \frac{(S + d)\ m}{(1 + t)\ sec} $$

However, I don't get the correct answer for my example above where $S = 19.595 \frac{m}{sec}, d = 0.788\ m, \textrm{and}\ t = 0.02\ sec$.

$$ S' = \frac{19.595 + 0.788}{1 + 0.2} = 19.983 $$ which is different from the $19.791$ that I computed above.

Can anyone please help?

Answer 1

There cannot be a function $f$ such that $S'=f(S,d,t).$ Suppose we do one experiment where we get $D=2,T=1,d=1$ and $t=1$. Then $S=2$ and $S'=3/2$. Now suppose we do a second experiment and get $D=4, T=2, d=1$ and $t=1$. Then $S=2$ and $S'=5/3$.

The values of $S,d,$ and $t$ are the same in the first and second experiments, but the value of $S'$ is different.

Answer 2

As you are obviously aware, if you have traveled a distance $D$ meters over the first $T$ seconds of traveling, and in the next $t$ seconds you travel an additional $d$ meters, one correct formula to find the speed over the entire travel time so far is

$$ \frac{D + d}{T + t} \text{ m/sec.} $$

This can also be viewed as a kind of weighted average of averages. If $S$ is your average speed during the first $T$ seconds and $s$ is your average speed during the next $t$ seconds, then $D = ST,$ $d = st,$ and $$ \frac{D + d}{T + t} = \frac{TS + ts}{T + s}. $$

The expression on the right side of that equation is a weighted average of the two speeds $S$ and $s,$ giving weight $T$ to $S$ and giving weight $t$ to $s.$

If all you know is $S,$ $d,$ and $t,$ you can compute $s,$ and you can suppose that $t$ is the weight of $s,$ but you have no way to guess the correct weight of $S.$ Your attempt was to assign weight $1$ to $S$, but as you can see, that doesn't work; in fact there is no single weight you can use that is any better than a blind guess.

The weight to assign to the speed $S$ is the piece of information you are missing when you try to derive the function $f(S,d,t).$

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The weight of the previous observations is also the reason why if you have a batting average of $.200$ when you step up to the plate and then you get a hit, your average will go up a lot more if this was your $6$th at-bat than if it was your $96$th at-bat. It isn't really all that different a problem from updating your speed.