A suitcase is placed on a horizontal conveyor belt. The coefficient...

For the suitcase not to slide on the conveyor belt, the maximum acceleration of the belt cannot exceed the maximum acceleration provided by static friction.

The maximum static friction force is given by:
$f_{s,\text{max}} = \mu_s m g$

The maximum acceleration $a_{\text{max}}$ of the suitcase without sliding is:
$a_{\text{max}} = \dfrac{f_{s,\text{max}}}{m} = \mu_s g$

Substituting the given values $\mu_s = 0.3$ and $g = 10 \text{ m/s}^2$:
$a_{\text{max}} = 0.3 \times 10 = 3 \text{ m/s}^2$

The belt starts from rest ($u = 0$) and accelerates uniformly to a final speed $v = 12 \text{ m/s}$. Using the first equation of motion $v = u + at$, the time taken is:
$t = \dfrac{v}{a}$

For the minimum time $t_{\text{min}}$, the acceleration must be maximum:
$t_{\text{min}} = \dfrac{12}{3} = 4.0 \text{ s}$

Answer: $4.0 \text{ s}$