[Q1] Describe the motion strategy that minimizes the travel time from A to B.

To minimize travel time, the train's average velocity must be maximized. This is achieved by reaching the highest possible speeds for the longest duration. The optimal strategy is a two-phase motion:

1. Maximum Acceleration: The train accelerates with its maximum possible acceleration, $a_1$.
2. Maximum Deceleration: It then immediately switches to decelerating with the maximum possible magnitude, $a_2$, timed such that it comes to a complete stop exactly at station B. Any coasting or use of less-than-maximum acceleration/deceleration would result in a lower average velocity and thus a longer travel time.

[Q2] What is the minimum travel time?

We can analyze the optimal motion using a velocity-time ($v-t$) graph. The journey consists of an acceleration phase followed by a deceleration phase, forming a triangle on the graph.

Let $t_1$ be the duration of acceleration and $t_2$ be the duration of deceleration. The total time is $T = t_1 + t_2$. Let $v_{max}$ be the maximum velocity achieved at the switch-over point.

From the definitions of acceleration and deceleration:

$$v_{max} = a_1 t_1 \implies t_1 = \frac{v_{max}}{a_1}$$ $$v_{max} = a_2 t_2 \implies t_2 = \frac{v_{max}}{a_2}$$

The total time $T$ can be expressed in terms of $v_{max}$:

$$T = t_1 + t_2 = \frac{v_{max}}{a_1} + \frac{v_{max}}{a_2} = v_{max} \left( \frac{1}{a_1} + \frac{1}{a_2} \right)$$

The total distance $s$ is the area under the triangular $v-t$ graph:

$$s = \frac{1}{2} (\text{base}) (\text{height}) = \frac{1}{2} T v_{max}$$

We can now solve these two equations for $T$ by eliminating $v_{max}$. From the distance equation, we find $v_{max} = \frac{2s}{T}$. Substituting this into the time equation:

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

Combining the terms in the parenthesis:

$$T^2 = 2s \left( \frac{a_2 + a_1}{a_1 a_2} \right)$$

Finally, solving for the minimum travel time $T$:

$$T = \sqrt{2s \frac{a_1+a_2}{a_1 a_2}}$$