The calculation for average speed has always been a misconception. The word average in calculating the average speed is not the same as the arithmetic mean or summing up the numbers and divides it by the number of terms. Below is the discussion of the formula for average speed as well as its derivation.

The average speed $(R_{ave})$  is equal to the total distance travelled by the object divided by the total travel time.

$R_{ave}=\displaystyle\frac{D_{total}}{T_{total}}$

This formula is totally different to our misconception that we can solve it by the following formula

$R_{ave}=\displaystyle\frac{R_1+R_2+\dots+R_n}{n}$

Here are some problems about average speed.

Problem 1:

Jane is riding her bike up the hill with a speed of 20 kph. After going up, she went down with an average speed of 30 kph. What is her average speed?

Solution:

Draft the path of object as follows.

Let $t_u$   be the time it takes for Jane goimg up.

Let $r_u$   be the rate of bike uphill.

Let $r_d$   be the rate downhill.

Let $t_d$   be the time it takes for Jane going down.

Let $d$   be the distance form starting point to finish point.

Uphill, we have the following equation

$d=r_u\cdot t_u$  (1)

Downhill, we have the following equation

$d=r_d\cdot t_d$  (2)

We use the same distance because the distance travelled downhill is equal to the distance travelled uphill.

To find the formula for average speed we have

$R_{ave}=\displaystyle\frac{D_{total}}{T_{total}}$

$R_{ave}=\displaystyle\frac{d_u+d_d}{t_u+t_d}$

But,  $d_u=d_d=d$

$R_{ave}=\displaystyle\frac{d+d}{t_u+t_d}$

$R_{ave}=\displaystyle\frac{2d}{t_u+t_d}$  (3)

From (1) and (2) respectively, we can rewrite them in terms of  $t_u,t_d$

$t_u=\displaystyle\frac{d}{r_u}$

$t_d=\displaystyle\frac{d}{r_d}$

Substitute the latter to (3)

$R_{ave}=\displaystyle\frac{d_u+d_d}{t_u+t_d}$

$R_{ave}=\displaystyle\frac{2d}{\frac{d}{r_u}+\frac{d}{r_d}}$

$R_{ave}=\displaystyle\frac{2d}{d(\frac{1}{r_u}+\frac{1}{r_d})}$

Cancel $d$

$R_{ave}=\displaystyle\frac{2}{\frac{1}{r_u}+\frac{1}{r_d}}$

This is already the formula that we need. We can also rewrite the formula in the following manner

$\displaystyle\frac{2}{R_{ave}}=\displaystyle\frac{1}{r_u}+\displaystyle\frac{1}{r_d}$

Or in general form,

$\displaystyle\frac{2}{R_{ave}}=\displaystyle\frac{1}{r_1}+\displaystyle\frac{1}{r_2}$

Going back to the formula we have $r_u=20 kph$ and $r_d=30 kph$

$R_{ave}=\displaystyle\frac{2}{\frac{1}{r_u}+\frac{1}{r_d}}$

$R_{ave}=\displaystyle\frac{2}{\frac{1}{20}+\frac{1}{30}}$

$R_{ave}=\displaystyle\frac{2\cdot 30\cdot 20}{30+20}$

$\boxed{R_{ave}=24 kph}$

Sample Problem 2:

Jake’s average speed in his entire trip from his house to Jam’s house and back home using his skateboard is $54$ meters per minute. He went to Jam’s house with a speed of $45$ meters per minute. What is his speed going back home?

Solution:

Since the formula has been derived, it would be very easy for us to solve for it.

Using the formula above we have,

$\displaystyle\frac{2}{R_{ave}}=\displaystyle\frac{1}{r_1}+\displaystyle\frac{1}{r_2}$

$\displaystyle\frac{2}{54}=\displaystyle\frac{1}{45}+\displaystyle\frac{1}{r_2}$

$\displaystyle\frac{1}{27}=\displaystyle\frac{1}{45}+\displaystyle\frac{1}{r_2}$

$\displaystyle\frac{1}{r_2}=\displaystyle\frac{1}{27}-\displaystyle\frac{1}{45}$

$\displaystyle\frac{1}{r_2}=\displaystyle\frac{2}{135}$

$\boxed{r_2=67.5 \frac{m}{min}}$