A train travels 2.97 km in 3 minutes. It accelerates uniformly, then travels at a constant speed, and finally retards uniformly to stop. The ratio of acceleration to retardation is 2:3. Find the maxim

Total time $T = 3 \text{ minutes} = 3 \times 60 \text{ s} = 180 \text{ s}$.
Total distance $D = 2.97 \text{ km} = 2.97 \times 1000 \text{ m} = 2970 \text{ m}$.
Maximum speed $v_{max} = 18 \text{ m/s}$.
Let $a$ be the acceleration and $r$ be the retardation.
The ratio of acceleration to retardation is $a:r = 2:3$, so $r = \frac{3}{2}a$.

The motion consists of three parts:

1. Uniform acceleration from 0 to $v_{max}$.
2. Constant velocity at $v_{max}$.
3. Uniform retardation from $v_{max}$ to 0.

Let $t_1$, $t_2$, $t_3$ be the times for acceleration, constant velocity, and retardation, respectively.
Let $s_1$, $s_2$, $s_3$ be the distances covered in these phases.

For acceleration phase:
Initial velocity $u=0$, final velocity $v=v_{max}=18 \text{ m/s}$.

The distance covered is the area of the triangle:

For retardation phase:
Initial velocity $u=v_{max}=18 \text{ m/s}$, final velocity $