3.49 A mail truck left $A$ for $B$. After 20 minutes, a second car left on the same route with a speed of 45 km/h. Catching up with the first car, the driver handed over a package and immediately returned at the same speed (the time spent on the stop and turn is negligible). At the moment when the first car arrived in $B$, the second car had only reached the midpoint of the way from the meeting point with the first car to point $A$. Find the speed of the first car, if the distance between $\boldsymbol{A}$ and $B$ is $40 \mathrm{kM}$.
3.49 Consider a coordinate system "path" ($s$ - in kilometers), "time" ($t$ - in hours). Let $AC$ (Fig. 3.12) be the graph of the motion of the first car with the desired speed $v = \tan \alpha$; $DE$ and $EF$ be the graph of the "there and back" motion of the second car with speed $\tan \beta = 45$; $AD = \frac{1}{3}$. It is known that $AB = 40$ and $G$ is the midpoint of the path $AH$. Let $AG = NK = y$. Then the time interval $DK = \frac{y}{\tan \beta} = \frac{y}{45}$. Geometrically, it is clear that $DK = KL = LM$, so the time interval of the first car's motion $AM = \frac{1}{3} + \frac{3y}{45}$, from which $\left(\frac{1}{3} + \frac{y}{15}\right) v = 40$.

The time interval $AL = \frac{1}{3} + \frac{2y}{45}$, $LE = AH = 2y$, so

$\left(\frac{1}{3} + \frac{2y}{45}\right) v = 2y$.

Solving the equations (1) and (2) simultaneously, we find $y = 15 \text{km}$ and $v = 30 \text{km/h}$.