To achieve the minimum travel time, the train must use its maximum acceleration $a_1$ for a duration $t_1$, immediately followed by its maximum deceleration $a_2$ for a duration $t_2$ until it comes to a stop. This strategy ensures the highest possible average speed.

Let $v_{max}$ be the maximum velocity reached at the end of the acceleration phase. The motion can be analyzed in two phases.

Phase 1: Acceleration The train accelerates from rest to $v_{max}$.

$$v_{max} = a_1 t_1 \implies t_1 = \frac{v_{max}}{a_1}$$

The distance covered during this phase is $S_1$.

$$v_{max}^2 = 2 a_1 S_1 \implies S_1 = \frac{v_{max}^2}{2 a_1}$$

Phase 2: Deceleration The train decelerates from $v_{max}$ to rest.

$$0 = v_{max} - a_2 t_2 \implies v_{max} = a_2 t_2 \implies t_2 = \frac{v_{max}}{a_2}$$

The distance covered during this phase is $S_2$.

$$0^2 = v_{max}^2 - 2 a_2 S_2 \implies S_2 = \frac{v_{max}^2}{2 a_2}$$

Total Time and Distance The total distance is $S = S_1 + S_2$.

$$S = \frac{v_{max}^2}{2 a_1} + \frac{v_{max}^2}{2 a_2} = \frac{v_{max}^2}{2} \left( \frac{1}{a_1} + \frac{1}{a_2} \right)$$

Solving for $v_{max}^2$:

$$v_{max}^2 = 2S \left( \frac{1}{\frac{1}{a_1} + \frac{1}{a_2}} \right) = 2S \frac{a_1 a_2}{a_1 + a_2}$$

The total minimum time is $T = t_1 + t_2$.

$$T = \frac{v_{max}}{a_1} + \frac{v_{max}}{a_2} = v_{max} \left( \frac{1}{a_1} + \frac{1}{a_2} \right)$$

Substitute the expression for $v_{max} = \sqrt{2S \frac{a_1 a_2}{a_1 + a_2}}$ into the equation for $T$:

$$T = \sqrt{2S \frac{a_1 a_2}{a_1 + a_2}} \left( \frac{a_2 + a_1}{a_1 a_2} \right)$$ $$T = \sqrt{2S} \frac{\sqrt{a_1 a_2}}{\sqrt{a_1 + a_2}} \frac{a_1 + a_2}{a_1 a_2} = \sqrt{2S} \frac{\sqrt{a_1 + a_2}}{\sqrt{a_1 a_2}}$$ $$T = \sqrt{2S \frac{a_1 + a_2}{a_1 a_2}}$$

This can also be written as:

$$T = \sqrt{2S \left(\frac{1}{a_1} + \frac{1}{a_2}\right)}$$