Calculate distance during acceleration.

The train starts from rest, so initial velocity $u = 0 \text{ m/s}$.
Acceleration $a = 2 \text{ m/s}^2$.
Time $t = 10 \text{ s}$.

First, find the final velocity ($v$) at the end of this phase, which will be the constant velocity for the next phase.

$$v = u + at$$ $$v = 0 \text{ m/s} + (2 \text{ m/s}^2)(10 \text{ s})$$ $$v = 20 \text{ m/s}$$

Now, calculate the distance traveled ($s_1$) during acceleration.

$$s_1 = ut + \frac{1}{2}at^2$$ $$s_1 = (0 \text{ m/s})(10 \text{ s}) + \frac{1}{2}(2 \text{ m/s}^2)(10 \text{ s})^2$$ $$s_1 = 0 + \frac{1}{2}(2)(100) \text{ m}$$ $$s_1 = 100 \text{ m}$$

The train moves with constant velocity $v = 20 \text{ m/s}$ (from Step 1).
Time $t = 20 \text{ s}$.

Calculate the distance traveled ($s_2$) during constant velocity.

$$s_2 = vt$$ $$s_2 = (20 \text{ m/s})(20 \text{ s})$$ $$s_2 = 400 \text{ m}$$

The train starts this phase with initial velocity $u = 20 \text{ m/s}$ (the constant velocity from Step 2).
It stops, so final velocity $v = 0 \text{ m/s}$.
Time $t = 5 \text{ s}$.

Calculate the distance traveled ($s_3$) during deceleration.

$$s_3 = \frac{u+v}{2}t$$ $$s_3 = \frac{20 \text{ m/s} + 0 \text{ m/s}}{2}(5 \text{ s})$$ $$s_3 = \frac{20}{2}(5) \text{ m}$$ $$s_3 = 10(5) \text{ m}$$ $$s_3 = 50 \text{ m}$$

Total distance = $s_1 + s_2 + s_3$
Total distance = $100 \text{ m} + 400 \text{ m} + 50 \text{ m}$
Total distance = $550 \text{ m}$

The total distance traveled is $550 \text{ m}$.

The correct option is B.

$$\boxed{\text{B. } 550 \text{ m}}$$