Answer
41.55 seconds
Work Step by Step
To calculate the total time, we must first figure out the time it takes to either accelerate to its max velocity of 5.08 m/s$^2$ or decelerate from that velocity to 0 (as they should both take the same amount of time).
$t_{1}=(5.08m/s)/(1.22m/s^2)=4.16s$
Since $t_{1}$ during acceleration is equal to $t_{2}$ during deceleration, we obtain:
$t_{1}+t_{2}=8.32s$
We know that together their distance would be $2(\Delta x)$
$2(10.58m)=21.16m$
The time for which the cab was at a constant velocity after accelerating (but before decelerating to a stop) is $190m-21.16m=168.84m$. We now calculate for the time when the cab was a constant velocity:
$t_{3}=(168.82m)/(5.08m/s)=33.23s$
Now, we can add up all times for our total time:
$t_{T}=8.32s+33.23s=41.55s$
