Answer
The total time is $~~3.0~s$
Work Step by Step
We can find the acceleration as the disk speeds up:
$v^2 = v_0^2+2ax$
$a = \frac{v^2-v_0^2}{2x}$
$a = \frac{(6.0~m/s)^2-0}{(2)(1.8~m)}$
$a = 10~m/s^2$
We can find the time of the acceleration period:
$v = v_0+at$
$t = \frac{v-v_0}{a}$
$t = \frac{6.0~m/s-0}{10~m/s^2}$
$t = 0.6~s$
We can find the time of the deceleration period:
$v = v_0+at$
$t = \frac{v-v_0}{a}$
$t = \frac{0-6.0~m/s}{-2.5~m/s^2}$
$t = 2.4~s$
The total time is $~~3.0~s$.
