Answer
The maximum speed is $~~36.3~m/s$
Work Step by Step
We can find the speed $v$ after the train accelerates over a distance of $550~m$:
$v^2 = v_0^2+2ax$
$v^2 = 0+2ax$
$v = \sqrt{2ax}$
$v = \sqrt{(2)(1.2~m/s^2)(550~m)}$
$v = 36.3~m/s$
The maximum speed is $~~36.3~m/s$.
