Answer
The total distance traveled by the disk is $~~9.0~m$
Work Step by Step
The distance traveled during the acceleration period is $1.8~m$
We can find the distance traveled during the deceleration period:
$v^2 = v_0^2+2ax$
$x = \frac{v^2-v_0^2}{2a}$
$x = \frac{0-(6.0~m/s)^2}{(2)(-2.5~m/s^2)}$
$x = 7.2~m$
The total distance traveled by the disk is $~~9.0~m$
