# 5. Solution:

Let $x$ units of distance/hour be the speed of the bus, $y$ units of distance/hour be the speed of the tractor, and $S$ be the length of the path AB. Then the speed of the truck is $-2y$ units of distance/hour. We can set up a system of equations and inequalities:

$$
\left\{\begin{array}{c}
\frac{s}{x}=5 \frac{5}{6} \\
\frac{S}{x+2 y} \geq 2.5 \\
\frac{S}{x+y}-\frac{S}{x+2 y} \geq 1
\end{array}\right.
$$

From the first equation, $S=5 \frac{5}{6} \cdot x=\frac{35}{6} \cdot x$. Substituting this result into the second inequality, we get: $\frac{35}{6} \cdot x \geq 2.5 \cdot x + 5 \cdot y$, from which $20 x \geq 30 y$ and $y \leq \frac{2}{3} x$.

Substituting $S$ into the second inequality, after transforming it: $S \cdot \frac{x+2 y-x-y}{(x+y) \cdot(x+2 y)} \geq 1 ; \quad \frac{35}{6} x y \geq x^{2} + 3 x y + 2 y^{2} ; \quad 6 x^{2} - 17 x y + 12 y^{2} \leq 0 ; \quad \frac{2}{3} \leq \frac{y}{x} \leq \frac{3}{4} ; \quad$ i.e., $y \geq \frac{2}{3} x$. From the two obtained estimates, it follows that $y \geq \frac{2}{3} x$. Let's find the time the tractor spends on the journey: $t=\frac{S}{y}=\frac{\frac{35}{6} x}{\frac{2}{3} x}=\frac{35}{4}=8$ hours 45 minutes. The required time is $9 + (8$ hours 45 minutes $)=17$ hours 45 minutes.
Answer: 17 hours 45 minutes.