2. First-grader Chukov runs one lap over rough terrain three minutes faster than his classmate Gekov (both run at a constant speed). If they start running simultaneously from one point of this lap but in opposite directions, they will not meet earlier than two minutes. If they start from the same point in the same direction, Chukov will not overtake Gekov on the lap later than 18 minutes. Determine the possible values for the time it takes Chukov to run one lap.
Solution. If Chukov runs a lap in $t$ minutes, then it will take Gekov $t+3$ minutes. If the length of the lap is $L$ meters, then Chukov's speed $V_{1}=\frac{L}{t}$, and Gekov's speed $\begin{aligned} V_{2}= & \frac{L}{t+3} . \text { According to the condition } \frac{L}{V_{1}+V_{2}} \geqslant 2, \frac{L}{V_{1}-V_{2}} \leqslant 18 . \\ & \quad \text { Therefore } \frac{L}{\frac{L}{t}+\frac{L}{t+3}} \geqslant 2, \text { hence } \frac{t(t+3)}{2 t+3} \geqslant 2, t^{2}+3 t \geqslant 4 t+6, t^{2}-t-6 \geqslant 0 . \text { The roots }\end{aligned}$ of the corresponding equation are: -2 and 3. Thus, $t \geqslant 3$.

Similarly, $\frac{L}{\frac{L}{t}-\frac{L}{t+3}} \leqslant 18$, hence $\frac{t(t+3)}{3} \leqslant 18, t^{2}+3 t-54 \leqslant 0$. The roots

of the corresponding equation are: -9 and 6. Thus, $t \leqslant 6$. Therefore, the answer is: $[3 ; 6]$.

Answer: From 3 to 6 minutes (including 3 and 6).

Grading criteria: 20 points - correct (not necessarily the same as above) solution and correct answer; $\mathbf{1 0}$ points - the solution is reduced to correct inequalities in terms of $t$, but further errors were made; $\mathbf{5}$ points - the equation and two inequalities in terms of poorly chosen unknown quantities are correctly written, but their solution is either not obtained or obtained with errors.