3.15 In the direction from $A$ to $B$, the car traveled for some time at a constant speed of $v_{1}=60 \mathrm{km} / \mathrm{h}$. The remaining part of the journey he covered in the same amount of time, but at a speed of $v_{2}=40 \mathrm{km} / \mathrm{h}$. In the opposite direction, the car traveled one half of the distance at a speed of $v_{3}=80 \mathrm{km} / \mathrm{h}$, and the other half at a speed of $v_{4}=45 \mathrm{km} / \mathrm{h}$. What is the average speed of the trip: a) from $A$ to $B$? b) from $B$ to $A$?
3.15 a) Since the car traveled at each of the specified speeds for equal time intervals, then

$$
v_{\mathrm{cp}}=\frac{v_{1}+v_{2}}{2}=\frac{60+40}{2}=50(\mathrm{KM} / \mathrm{4})
$$

b) The return trip consists of two equal segments of the path (let's assume each is $s$ km), which were covered by the car in different time intervals; therefore, it would be incorrect to assume that

$$
v_{\mathrm{cp}}=\frac{v_{3}+v_{4}}{2}=\frac{80+45}{2}=62.5(\mathrm{KM} / \mathrm{4})
$$

Let the car travel $x$ hours at speed $v_{3}$ and $y$ hours at speed $v_{4}$. Then $v_{3} x=v_{4} y=s$, from which $x=\frac{v_{4} y}{v_{3}}$. Therefore, the average speed

$$
v_{\mathrm{cp}}=\frac{2 s}{x+y}=\frac{2 v_{4} y}{\frac{v_{4} y}{v_{3}}+y}=\frac{2 v_{3} v_{4}}{v_{3}+v_{4}}=\frac{2 \cdot 80 \cdot 45}{125}=57.6(\mathrm{~K} / \mathrm{u})
$$

Answer: a) 50 km/h; b) 57.6 km/h.