During the first half of the travel time, a car moves at v_1…

During the first half of the travel time, a car moves at $v_1 = 54\ \mathrm{km/h}$ along a straight segment at an angle $\alpha_1 = 45^\circ$ to a reference direction. During the second half of the time, it moves at $v_2 = 72\ \mathrm{km/h}$ along a straight segment at $\alpha_2 = 60^\circ$ to the same direction. What is the average speed of the car?

Explanation of Method

Two equal time intervals on a bent path.

Step 1: Average speed ignores direction

As in Translational Kinematics 026, average speed is path length divided by time, so the angles do not matter. Let the total time be $T$: $s_1 = v_1 \frac{T}{2}$.

\[ s_2 = v_2 \frac{T}{2}. \]

Step 2: Equal halves of the time

\[ v_{\text{avg}} = \frac{s_1 + s_2}{T} = \frac{\frac{T}{2}(v_1 + v_2)}{T} = \frac{v_1 + v_2}{2}. \]

This is the arithmetic mean from Translational Kinematics 001.

Step 3: Substitute the values

\[ v_{\text{avg}} = \frac{54 + 72}{2} = \frac{126}{2}. \]

\[v_{\text{avg}} = 63\ \mathrm{km/h}\]

Check with a concrete time

$ T = 2\ \mathrm{h}: \quad s_1 = 54\ \mathrm{km}, \quad s_2 = 72\ \mathrm{km} $

$ v_{\text{avg}} = \dfrac{126\ \mathrm{km}}{2\ \mathrm{h}} = 63\ \mathrm{km/h}\ \checkmark $

$45^\circ$ $60^\circ$ $v_1 = 54\ \mathrm{km/h}$ $v_2 = 72\ \mathrm{km/h}$ $\text{displacement}$ $\text{reference}$

first part
second part
displacement

Both parts take the same time (3 equal time steps each), so the faster second part is longer. Average speed uses the total length of the path; average velocity would use the dashed displacement.

Result

\[\boxed{\,v_{\text{avg}} = 63\ \mathrm{km/h}\,}\]

Remark

Where the angles would matter. The average velocity uses the displacement. The two segments differ in direction by $15^\circ$, so its magnitude is $\tfrac12\sqrt{v_1^2 + v_2^2 + 2v_1v_2\cos 15^\circ} \approx 62.5\ \mathrm{km/h}$, directed at about $53.6^\circ$ to the reference direction. It is only slightly less than the average speed because the path bends only a little.