Answer: In 40 minutes.

Solution: The diagonals of a rectangle have the same length and are longer than any of its sides. In one hour, together the two cars would travel three times the length of side $B C$ and three times the diagonal, since one car travels two sides equal to $B C$ and one diagonal in an hour, while the other car travels two diagonals and one side equal to $B C$ in an hour. Therefore, in one-third of an hour, or 20 minutes, the two cars together will travel one side equal to $B C$ and one equal to the diagonal. The entire described route to the meeting point on $B D$ they will cover in twice this time.

Comment: Correct answer without justification - 1 point.

Correct answer obtained by example - 3 points.

Note: It can be proven (but is not required in this problem) that the meeting will definitely occur on the road $B D$.
7.2. Points $A, B, C, D$ are located at the vertices of rectangle $ABCD$, its sides and diagonals $AC$ and $BD$ are roads. The first car traveled the route $B \rightarrow$ $C \rightarrow A \rightarrow D$ in one hour, and the second car traveled the route $D \rightarrow B \rightarrow C \rightarrow A$ in one hour. After how much time will the cars meet if they simultaneously leave point $C$: the first one on the route $C \rightarrow$ $B \rightarrow D$, and the second one on the route $C \rightarrow A \rightarrow D \rightarrow B$, and the meeting will occur on the road $BD$? (The speeds of both cars are constant).