Given displacement function:

To solve this problem, we first need to find the expressions for velocity and acceleration by differentiating the displacement function.

Given displacement function: $S(t) = t^4 - 2t^2 - 5t^3$ We can reorder it for clarity: $S(t) = t^4 - 5t^3 - 2t^2$

Velocity is the first derivative of displacement with respect to time: $V(t) = \frac{dS}{dt} = \frac{d}{dt}(t^4 - 5t^3 - 2t^2) = 4t^3 - 15t^2 - 4t$

Acceleration is the first derivative of velocity with respect to time (or the second derivative of displacement): $A(t) = \frac{dV}{dt} = \frac{d}{dt}(4t^3 - 15t^2 - 4t) = 12t^2 - 30t - 4$

Now we can solve each part of the question.

A) Determine when $t=3$:

Step 1: Calculate displacement at $t=3$. Substitute $t=3$ into the displacement function $S(t)$. $S(3) = (3)^4 - 5(3)^3 - 2(3)^2$ $S(3) = 81 - 5(27) - 2(9)$ $S(3) = 81 - 135 - 18$ $S(3) = -72 \text{ m}$ The displacement at $t=3$ is $\boxed{-72 m}$.

Step 2: Calculate velocity at $t=3$. Substitute $t=3$ into the velocity function $V(t)$. $V(3) = 4(3)^3 - 15(3)^2 - 4(3)$ $V(3) = 4(27) - 15(9) - 12$ $V(3) = 108 - 135 - 12$ $V(3) = -39 \text{ m/s}$ The velocity at $t=3$ is $\boxed{-39 m/s}$.

Step 3: Calculate acceleration at $t=3$. Substitute $t=3$ into the acceleration function $A(t)$. $A(3) = 12(3)^2 - 30(3) - 4$ $A(3) = 12(9) - 90 - 4$ $A(3) = 108 - 90 - 4$ $A(3) = 14 m/s^2$ The acceleration at $t=3$ is $\boxed{14 m/s^2}$.

B) Time when velocity is 0:

Step 4: Set the velocity function $V(t)$ to 0 and solve for $t$. $V(t) = 4t^3 - 15t^2 - 4t = 0$ Factor out $t$: $t(4t^2 - 15t - 4) = 0$ This gives two possibilities: $t=0$ or $4t^2 - 15t - 4 = 0$. Solve the quadratic equation $4t^2 - 15t - 4 = 0$. We can factor it: $(4t+1)(t-4) = 0$ This yields two more solutions: $4t+1 = 0 \Rightarrow t = -\frac{1}{4} \text{ s}$ $t-4 = 0 \Rightarrow t = 4 \text{ s}$ Since time cannot be negative in this physical context, the times when velocity is 0 are $t=0$ s and $t=4$ s. The times when velocity is 0 are $\boxed{0 s and 4 s}$.

C) Acceleration at $t=2$:

Step 5: Calculate acceleration at $t=2$. Substitute $t=2$ into the acceleration function $A(t)$. $A(2) = 12(2)^2 - 30(2) - 4$ $A(2) = 12(4) - 60 - 4$ $A(2) = 48 - 60 - 4$ $A(2) = -16 m/s^2$ The acceleration at $t=2$ is $\boxed{-16 m/s^2}$.