447. A motorcyclist left point A at a speed of 45 km/h. After 40 minutes, a car left A in the same direction at a speed of 60 km/h. How much time after the car's departure will the distance between it and the motorcyclist be 36 km?
$\triangle$ Important question: at the moment when the car is 36 km away from the motorcycle, will it be ahead or behind the motorcycle?

In 40 minutes, the motorcycle will travel a distance of $45 \cdot \frac{2}{3}$ km $=30$ km, which is less than 36 km. Therefore, at the moment the car departs, it is 30 km behind the motorcycle, and the distance between them will continue to decrease until the car overtakes the motorcycle. This means that the car will be 36 km ahead of the motorcycle after the overtaking.

![](https://cdn.mathpix.com/cropped/2024_05_21_88152b3867f80e3be9a3g-155.jpg?height=94&width=1106&top_left_y=1541&top_left_x=475)

Fig. 1

In Fig. 1, points A and B denote the locations of the car and the motorcycle, respectively, at the moment the car departs, and points $\mathrm{A}_{1}$ and $\mathrm{B}_{1}$ denote their locations at the moment the car is 36 km ahead of the motorcycle. Then $\mathrm{AB}=30$ km, $\mathrm{B}_{1} \mathrm{~A}_{1}=36$ km.

Let the unknown be the distance $\mathrm{BB}_{1}: \mathrm{BB}_{1}=x$ km. If we find this distance, we will also find the time the car takes to travel the path $\mathrm{AA}_{1}$.

The verbal "equation" here is: the time the car takes to travel the path $\mathrm{AA}_{1}$ equals the time the motorcycle takes to travel the path $\mathrm{BB}_{1}$. Express both times in terms of $x$ and set the expressions equal:

$$
\frac{30+x+36}{60}=\frac{x}{45}
$$

Solve this equation:

$$
3(66+x)=4 x, \quad 198+3 x=4 x, \quad x=198
$$

From this, the time the car takes to travel the path $\mathrm{AA}_{1}$ is $\frac{198}{45}=\frac{22}{5}=4 \frac{2}{5}$ (hours), i.e., 4 hours 24 minutes. Answer: 4 hours 24 minutes.