Answer
After 2.22 seconds, the box will fall off the truck.
The truck travels a distance of 5.42 meters.
Work Step by Step
We can use $\mu_s$ to find the maximum possible acceleration of the box from the force of static friction.
$mg~\mu_s = ma$
$a = g~\mu_s = (9.80~m/s^2)(0.19) = 1.86~m/s^2$
Since the maximum possible acceleration of the box is less than the truck's acceleration, the box will start to slide relative to the truck. The force of kinetic friction will cause the box to accelerate in the same direction as the truck's acceleration.
$F_f = ma$
$mg~\mu_k = ma$
$a = g~\mu_k = (9.80~m/s^2)(0.15) = 1.47~m/s^2$
In time t:
The distance moved by the truck is $\frac{1}{2}(2.20~m/s^2)t^2$.
The distance moved by the box is $\frac{1}{2}(1.47~m/s^2)t^2$.
We need the difference between the distances to be 1.80 meters.
$\frac{1}{2}(2.20~m/s^2)t^2 - \frac{1}{2}(1.47~m/s^2)t^2 = 1.80~m$
$t^2 = \frac{3.60~m}{2.20~m/s^2-1.47~m/s^2}$
$t = 2.22~s$
After 2.22 seconds, the box will fall off the truck.
We can find the distance that the truck travels.
$x = \frac{1}{2}(2.20~m/s^2)(2.22~s)^2$
$x = 5.42~m$
The truck travels a distance of 5.42 meters.
