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

For the acceleration phase:
Initial velocity, $u = 0$
Acceleration, $a_1 = 2.0 \text{ m/s}^2$
Time, $t_1 = 0.5 \text{ min} = 30 \text{ s}$

Maximum velocity reached, $v = u + a_1 t_1 = 0 + 2.0 \times 30 = 60 \text{ m/s}$

Distance covered during acceleration, $s_1 = \dfrac{1}{2} a_1 t_1^2 = \dfrac{1}{2} \times 2.0 \times (30)^2 = 900 \text{ m}$

For the deceleration phase:
Initial velocity, $v = 60 \text{ m/s}$
Final velocity, $v_f = 0$
Time, $t_2 = 1 \text{ min} = 60 \text{ s}$

Distance covered during deceleration, $s_2 = \dfrac{v + v_f}{2} t_2 = \dfrac{60 + 0}{2} \times 60 = 1800 \text{ m}$

Total distance moved by the train, $s = s_1 + s_2 = 900 + 1800 = 2700 \text{ m}$

Converting to kilometers, $s = 2.7 \text{ km}$