On the first half of its path, a straight segment at an…
On the first half of its path, a straight segment at an angle $\alpha_1 = 60^\circ$ to a reference direction, a car moves at $v_1 = 72\ \mathrm{km/h}$. On the second half, a straight segment at $\alpha_2 = 30^\circ$ to the same direction, it moves at $v_2 = 36\ \mathrm{km/h}$. What is the average speed of the car?
Explanation of Method
Two equal parts of a bent path.
Step 1: Average speed ignores direction
Average speed is the length of the path divided by the time. The directions of the segments do not change their lengths or the times, so the angles play no role here. Let the path length be $L$:
\[ t_1 = \frac{L/2}{v_1} = \frac{L}{2v_1}, \]
\[ t_2 = \frac{L/2}{v_2} = \frac{L}{2v_2}. \]
Step 2: Equal halves of the path
\[ v_{\text{avg}} = \frac{L}{t_1 + t_2} = \frac{L}{\dfrac{L}{2v_1} + \dfrac{L}{2v_2}} = \frac{2 v_1 v_2}{v_1 + v_2}. \]
This is the harmonic mean from Translational Kinematics 002.
Step 3: Substitute the values
\[ v_{\text{avg}} = \frac{2 \cdot 72 \cdot 36}{72 + 36} = \frac{5184}{108}. \]
\[v_{\text{avg}} = 48\ \mathrm{km/h}\]
Check with a concrete length
$ L = 72\ \mathrm{km}: \quad t_1 = \dfrac{36}{72} = 0.5\ \mathrm{h}, \quad t_2 = \dfrac{36}{36} = 1\ \mathrm{h} $
$ v_{\text{avg}} = \dfrac{72\ \mathrm{km}}{1.5\ \mathrm{h}} = 48\ \mathrm{km/h}\ \checkmark $
- first part
- second part
- displacement
Result
\[\boxed{\,v_{\text{avg}} = 48\ \mathrm{km/h}\,}\]
Remark
Where the angles would matter. For the average velocity (displacement divided by time) the directions count. The two equal halves meet at an angle of $60^\circ - 30^\circ = 30^\circ$, so the displacement is $L\cos 15^\circ \approx 0.966\,L$, pointing at $45^\circ$ to the reference direction. The average velocity is then $0.966 \cdot 48 \approx 46.4\ \mathrm{km/h}$ at $45^\circ$.