Problem 4. A car traveled the path from place $A$ to place $B$ at a speed of $60 \mathrm{~km} / \mathrm{h}$, and the path from place $B$ to place $A$ at a speed of $40 \mathrm{~km} / \mathrm{h}$. Determine the average speed of the car's movement.
Solution. Let's denote the distance from $A$ to $B$ by $s$, the speed at which the car travels from $A$ to $B$ by $v_{1}$, and the speed at which it travels from $B$ to $A$ by $v_{2}$. Then the corresponding times are $t_{1}=\frac{s}{v_{1}}$ and $t_{2}=\frac{s}{v_{2}}$. Therefore, if $v_{S}$ is the average speed at which the car travels the entire distance over time $t_{1}+t_{2}$, then

$$
v_{S}=\frac{2 s}{t_{1}+t_{2}}=\frac{2 s}{\frac{s}{v_{1}}+\frac{s}{v_{2}}}=\frac{2}{\frac{1}{v_{1}}+\frac{1}{v_{2}}}=\frac{2 v_{1} v_{2}}{v_{1}+v_{2}}=\frac{2 \cdot 60 \cdot 40}{60+40}=48 \mathrm{~km} / \mathrm{h}
$$