Answer
The total travel time is $~~60.6~s$
Work Step by Step
We can find the time $t$ to travel $550~m$ during the acceleration period:
$x = \frac{1}{2}at^2$
$t = \sqrt{\frac{2x}{a}}$
$t = \sqrt{\frac{(2)(550~m)}{1.2~m/s^2}}$
$t = 30.28~s$
Note that the rate of acceleration is $1.2~m/s^2$ and the rate of deceleration is $-1.2~m/s^2$
By symmetry, the time to travel $550~m$ during the deceleration period is also $t = 30.28~s$
The total travel time is $~~60.6~s$.
