On the first third of the total distance, a car moves at…

On the first third of the total distance, a car moves at speed $v_1$. On the remaining two thirds, it moves at $v_2 = 54\ \mathrm{km/h}$. The average speed over the entire journey is $\langle v \rangle = 36\ \mathrm{km/h}$. What is the speed $v_1$?

Step 1: Split the distance

Let the total distance be $s$: $s_1 = \frac{s}{3}, \qquad s_2 = \frac{2s}{3}$.

Step 2: Time on each part

\[ t_1 = \frac{s_1}{v_1} = \frac{s}{3v_1}, \qquad t_2 = \frac{s_2}{v_2} = \frac{2s}{3v_2}. \]

Step 3: Write the average speed

\[ \langle v \rangle = \frac{s}{t_1 + t_2} = \frac{s}{\dfrac{s}{3v_1} + \dfrac{2s}{3v_2}} = \frac{3}{\dfrac{1}{v_1} + \dfrac{2}{v_2}}. \]

The distance $s$ cancels again.

Step 4: Solve for $v_1$

Flip both sides and isolate $\dfrac{1}{v_1}$: \[\frac{3}{\langle v \rangle} = \frac{1}{v_1} + \frac{2}{v_2} \quad\Longrightarrow\quad \frac{1}{v_1} = \frac{3}{\langle v \rangle} - \frac{2}{v_2} = \frac{3v_2 - 2\langle v \rangle}{\langle v \rangle\, v_2}.\]

\[ v_1 = \frac{\langle v \rangle\, v_2}{3v_2 - 2\langle v \rangle}. \]

Step 5: Substitute the values

\[ v_1 = \frac{36 \cdot 54}{3\cdot 54 - 2\cdot 36}\ \mathrm{km/h} = \frac{1944}{162 - 72}\ \mathrm{km/h} = \frac{1944}{90}\ \mathrm{km/h}. \]

\[v_1 = 21.6\ \mathrm{km/h} = 6\ \mathrm{m/s}\]

Check with a concrete distance

Take $s = 108\ \mathrm{km}$, so $s_1 = 36\ \mathrm{km}$ and $s_2 = 72\ \mathrm{km}$:

$ t_1 = \dfrac{36\ \mathrm{km}}{21.6\ \mathrm{km/h}} = 1\tfrac{2}{3}\ \mathrm{h} = 1\ \mathrm{h}\ 40\ \mathrm{min} $

$ t_2 = \dfrac{72\ \mathrm{km}}{54\ \mathrm{km/h}} = 1\tfrac{1}{3}\ \mathrm{h} = 1\ \mathrm{h}\ 20\ \mathrm{min} $

$ \langle v \rangle = \dfrac{108\ \mathrm{km}}{3\ \mathrm{h}} = 36\ \mathrm{km/h}\ \checkmark $