Why is the average rate isn't the average of the two rates given for the two legs of the journey?

Question

If an object moves the same distance twice, but at different rates, then the average rate will never be an average of the two rates given for the two legs of the journey.

Why is that so?

Answer 1

It would only be the average if the object is moving for an equal amount of time for both the higher speed and the lower speed, but that is not the case, because the object is moving for a longer time period at the lower speed.

Answer 2

Computationally: let the distance be $d$ and the times taken to travel it $t_1$ and $t_2$. The average rate is $\frac{2d}{t_1+t_2}$, while the average of the rates is $\frac12(\frac d{t_1}+\frac d{t_2}) = \frac{(t_1+t_2)d}{2t_1t_2}$. To compare these, we subtract the average rate from the average of the rates:

$$ \begin{align} \frac{(t_1+t_2)d}{2t_1t_2} - \frac{2d}{t_1+t_2} &= \frac{(t_1+t_2)^2d-4dt_1t_2}{2t_1t_2(t_1+t_2)} \\&= \frac{(t_1-t_2)^2d}{2t_1t_2(t_1+t_2)}\end{align} $$ This is never negative, and as long as $t_1 \neq t_2$ (i.e., the two rates are not the same), it is positive, so the average of the rates is greater than the average rate.

Intuitively: The average of rates weights the two rates equally. Taking the average rate over the entire trip takes the average over time, so it gives more weight to the rate which lasts the longer time. Some thought makes it clear that if the distance is the same, there will be more time traveled at the slower rate than the faster weight, so the slower rate is given more weight, resulting in a lower average.

Answer 3

The average rate is the rate necessary for the object to complete the distance in the net time. By this definition, there is no reason other than false intuition to believe that the average rate is the mean of the two given rates.

If the distance for both rates is $d$ and we have $t_1$, and $t_2$ to be the time taken for the duration of the first and second rate respectively, then $\dots$

$$r_1 = \frac{d}{t_1} $$ $$r_2 = \frac{d}{t_2} $$ $$r_{average} = \frac{2d}{t_1 + t_2}$$

There is no reason why the average rate and average of the rates must be the same or different.