During the first half of the total travel time, a car moves…
During the first half of the total travel time, a car moves at a speed of $v_1 = 54\ \mathrm{km/h}$. During the second half of the travel time, it moves at a speed of $v_2 = 36\ \mathrm{km/h}$. What is the average speed of the car?
Explanation of Method
Two equal halves of the travel time.
Step 1: Definition of average speed
Average speed is the total distance divided by the total time:
\[ v_{\text{avg}} = \frac{s}{t} = \frac{s_1 + s_2}{t}. \]
Let the total travel time be $t$. Each half then lasts $\dfrac{t}{2}$.
Step 2: Distance covered in each half
\[ s_1 = v_1 \cdot \frac{t}{2}, \qquad s_2 = v_2 \cdot \frac{t}{2}. \]
On a speed–time graph, each distance is the area of a rectangle: its width is the time interval and its height is the speed.
- distance in 1st half
- distance in 2nd half
- average speed
Step 3: General formula for equal time intervals
\[ v_{\text{avg}} = \frac{v_1\frac{t}{2} + v_2\frac{t}{2}}{t} = \frac{\frac{t}{2}\,(v_1 + v_2)}{t} = \frac{v_1 + v_2}{2}. \]
The total time $t$ cancels. When the time intervals are equal, the average speed is the arithmetic mean of the two speeds.
Step 4: Substitute the values
\[ v_{\text{avg}} = \frac{54\ \mathrm{km/h} + 36\ \mathrm{km/h}}{2} = \frac{90\ \mathrm{km/h}}{2}. \]
\[v_{\text{avg}} = 45\ \mathrm{km/h} = 12.5\ \mathrm{m/s}\]
Check with a concrete time
Suppose the trip lasts $t = 2\ \mathrm{h}$. Then:
$ s_1 = 54\ \mathrm{km/h}\cdot 1\ \mathrm{h} = 54\ \mathrm{km} $
$ s_2 = 36\ \mathrm{km/h}\cdot 1\ \mathrm{h} = 36\ \mathrm{km} $
$ v_{\text{avg}} = \dfrac{54\ \mathrm{km} + 36\ \mathrm{km}}{2\ \mathrm{h}} = \dfrac{90\ \mathrm{km}}{2\ \mathrm{h}} = 45\ \mathrm{km/h}\ \checkmark $
Result
\[\boxed{\,v_{\text{avg}} = 45\ \mathrm{km/h} = 12.5\ \mathrm{m/s}\,}\]
Remark
Common pitfall. The formula $v_{\text{avg}} = \dfrac{2v_1v_2}{v_1+v_2}$ (the harmonic mean) applies when the distance is split into two equal halves, not the time. In that case the car spends more time at the slower speed, so the average speed is lower:
| Split into equal… | Formula | Result |
|---|---|---|
| time (this problem) | $\dfrac{v_1+v_2}{2}$ | $45\ \mathrm{km/h}$ |
| distance | $\dfrac{2v_1v_2}{v_1+v_2}$ | $43.2\ \mathrm{km/h}$ |
Because this problem says half of the travel time, the correct answer is $45\ \mathrm{km/h}$. The equal-distance case is worked out in Translational Kinematics 002.