Starting from rest, a train travels with a constant acceleration of...

Maximum speed attained by the train is v_\max = u + a_1 t_1 = 0 + 2.0 \times 30 = 60 \text{ m/s}.

Half of the maximum speed is $v = \dfrac{60}{2} = 30 \text{ m/s}$.

The train attains this speed twice: once during acceleration and once during deceleration.

During the acceleration phase, the position $s_1$ at $v = 30 \text{ m/s}$ is given by:

$v^2 = u^2 + 2a_1 s_1$

$30^2 = 0 + 2(2.0)s_1 \Rightarrow s_1 = \dfrac{900}{4} = 225 \text{ m}$

The total distance covered during the acceleration phase is:

$S_1 = \dfrac{1}{2} a_1 t_1^2 = \dfrac{1}{2}(2.0)(30)^2 = 900 \text{ m}$

During the deceleration phase, the train comes to rest from $60 \text{ m/s}$ in $60 \text{ s}$. The acceleration is:

$a_2 = \dfrac{0 - 60}{60} = -1.0 \text{ m/s}^2$

The distance $s_2$ covered during deceleration to reach $v = 30 \text{ m/s}$ is given by:

v^2 = v_\max^2 + 2a_2 s_2

$30^2 = 60^2 + 2(-1.0)s_2 \Rightarrow 2s_2 = 3600 - 900 = 2700 \Rightarrow s_2 = 1350 \text{ m}$

The position of the train from the starting point during deceleration is:

$S_{total} = S_1 + s_2 = 900 + 1350 = 2250 \text{ m} = 2.25 \text{ km}$