448. A cyclist set off from point A to point B, and 15 minutes later, a car set off after him. Halfway from A to B, the car caught up with the cyclist. When the car arrived at B, the cyclist still had to cover another third of the entire distance. How long will it take the cyclist to travel the distance from A to B?
$\triangle$ Let's take the path AB as a unit. (This can be done in cases where there is no distance given in linear units in the problem data.) Let's denote the speed of the cyclist by $x$ (in fractions of the path per hour).

During the time it took the car to travel the second half of the path, the cyclist traveled $\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$ of the path AB. Therefore, the car's speed is $\frac{1}{2}: \frac{1}{6}=3$ times greater than the cyclist's speed, i.e., it is $3x$.

The verbal "equation" here is: the time it took the cyclist to travel the first half of the path AB is 15 minutes $=\frac{1}{4}$ hour more than the time it took the car to travel the same half. We get:

$$
\frac{1}{2 x}-\frac{1}{2 \cdot 3 x}=\frac{1}{4}
$$

Solving this equation, we find $x: x=\frac{4}{3}$. Then the time it took the cyclist to travel the path AB is $1: \frac{4}{3}=\frac{3}{4}$ hours, i.e., 45 minutes.

Answer: 45 minutes.