The speed of a car during the first half of the distance is…

The speed of a car during the first half of the distance is $v_1 = 36\ \mathrm{km/h}$, and during the second half of the distance it is $v_2 = 54\ \mathrm{km/h}$. What is the average speed of the car over the entire journey?

Explanation of Method

Two equal halves of the distance.

Step 1: Time spent on each half

Let the total distance be $s$. Each half is $\dfrac{s}{2}$ long, and time is distance divided by speed: \[t_1 = \frac{s/2}{v_1} = \frac{s}{2v_1}, \qquad t_2 = \frac{s/2}{v_2} = \frac{s}{2v_2}.\]

This time the distances are equal, so the times are not. The slower half takes longer.

$v\ (\mathrm{km/h})$ $t$ $0$ $t_1$ $t_1+t_2$ $54$ $43.2$ $36$ $v_1 = 36\ \mathrm{km/h}$ $v_2 = 54\ \mathrm{km/h}$ $s_1 = \tfrac{s}{2}$ $s_2 = \tfrac{s}{2}$ $v_{\text{avg}} = 43.2\ \mathrm{km/h}$

1st half of distance
2nd half of distance
average speed

Both rectangles have the same area (equal distances), so the slower rectangle must be wider: the car spends more time at 36 km/h than at 54 km/h (here 1.5 h vs 1 h for $s = 108\ \mathrm{km}$). That extra time pulls the average below the midpoint 45 km/h.

Step 2: General formula for equal distances

\[ v_{\text{avg}} = \frac{s}{t_1 + t_2} = \frac{s}{\dfrac{s}{2v_1} + \dfrac{s}{2v_2}} = \frac{2}{\dfrac{1}{v_1} + \dfrac{1}{v_2}} = \frac{2 v_1 v_2}{v_1 + v_2}. \]

The distance $s$ cancels. This expression is the harmonic mean of the two speeds.

Step 3: Substitute the values

\[ v_{\text{avg}} = \frac{2 \cdot 36 \cdot 54}{36 + 54} = \frac{3888}{90}. \]

\[v_{\text{avg}} = 43.2\ \mathrm{km/h} = 12\ \mathrm{m/s}\]

Check with a concrete distance

Take $s = 108\ \mathrm{km}$, so each half is $54\ \mathrm{km}$:

$ t_1 = \dfrac{54\ \mathrm{km}}{36\ \mathrm{km/h}} = 1.5\ \mathrm{h} $

$ t_2 = \dfrac{54\ \mathrm{km}}{54\ \mathrm{km/h}} = 1\ \mathrm{h} $

$ v_{\text{avg}} = \dfrac{108\ \mathrm{km}}{2.5\ \mathrm{h}} = 43.2\ \mathrm{km/h}\ \checkmark $

Result

\[\boxed{\,v_{\text{avg}} = 43.2\ \mathrm{km/h} = 12\ \mathrm{m/s}\,}\]

Remark

Compare with Translational Kinematics 001. The two speeds are the same 36 and 54 km/h, but the answer is lower (43.2 instead of 45 km/h). With equal distances, the car spends more of its time driving slowly, so the slower speed weighs more in the average.