#### Answer

The distance between the stop signs is 216 meters.

#### Work Step by Step

We can divide the motion into three distances $x_1$, $x_2$, and $x_3$;
$x_1 = \frac{1}{2}at^2$
$x_1 = \frac{1}{2}(4.0~m/s^2)(6.0~s)^2$
$x_1 = 72~m$
After 6.0 seconds, the speed is $(4.0~m/s^2)(6.0~s)$ which is 24 m/s.
Therefore;
$x_2 = v~t = (24~m/s)(2.0~s) = 48~m$
$x_3 = \frac{v^2-v_0^2}{2a}$
$x_3 = \frac{0-(24~m/s)^2}{(2)(-3.0~m/s^2)}$
$x_3 = 96~m$
We can then find the total distance $x$ between the stop signs:
$x = x_1+x_2+x_3$
$x = 72~m+48~m+96~m$
$x = 216~m$
The distance between the stop signs is 216 meters.