Here's the solution for the object's displacement and average velocity: The position of the particle is given by $y(t) = (t^3 - 4t^2 - 6t)\text{m}$. We need to find the displacement and average velocity for the time interval from $t = 3\text{s}$ to $t = 5\text{s}$. Step 1: Calculate the position at $t = 3\text{s}$. Substitute $t=3$ into the position function: $$y(3) = (3^3 - 4(3^2) - 6(3))\text{m}$$ $$y(3) = (27 - 4(9) - 18)\text{m}$$ $$y(3) = (27 - 36 - 18)\text{m}$$ $$y(3) = (-9 - 18)\text{m}$$ $$y(3) = -27\text{m}$$ Step 2: Calculate the position at $t = 5\text{s}$. Substitute $t=5$ into the position function: $$y(5) = (5^3 - 4(5^2) - 6(5))\text{m}$$ $$y(5) = (125 - 4(25) - 30)\text{m}$$ $$y(5) = (125 - 100 - 30)\text{m}$$ $$y(5) = (25 - 30)\text{m}$$ $$y(5) = -5\text{m}$$ Step 3: Calculate the displacement ($\Delta y$). Displacement is the change in position, $\Delta y = y(t_2) - y(t_1)$. $$\Delta y = y(5) - y(3)$$ $$\Delta y = (-5\text{m}) - (-27\text{m})$$ $$\Delta y = -5\text{m} + 27\text{m}$$ $$\Delta y = 22\text{m}$$ The object's displacement is $\boxed{\text{22 m}}$. Step 4: Calculate the average velocity ($v_{avg}$). Average velocity is the displacement divided by the time interval, $v_{avg} = \frac{\Delta y}{\Delta t}$. The time interval is $\Delta t = 5\text{s} - 3\text{s} = 2\text{s}$. $$v_{avg} = \frac{22\text{m}}{2\text{s}}$$ $$v_{avg} = 11\text{m/s}$$ The object's average velocity is $\boxed{\text{11 m/s}}$.