The problem involves the equation of motion of a particle given by:

$s(t) = t^3 - 12t$

where $s$ is the displacement in meters, and $t$ is time in seconds.

(a) Find the velocity and acceleration as functions of $t$.

1. Velocity, $v(t)$, is the first derivative of the position function $s(t)$ with respect to time $t$: $v(t) = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 12t) = 3t^2 - 12$

2. Acceleration, $a(t)$, is the first derivative of the velocity function, or the second derivative of the position function: $a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12) = 6t$

Thus, the velocity and acceleration as functions of $t$ are: $v(t) = 3t^2 - 12, \quad a(t) = 6t$

(b) Find the acceleration, in m/s², after 4 seconds.

To find the acceleration at $t = 4$ seconds, we substitute $t = 4$ into the acceleration function $a(t) = 6t$: $a(4) = 6(4) = 24 \, \text{m/s}^2$

(c) Find the acceleration, in m/s², when the velocity is 0.

To find the acceleration when the velocity is zero, we first set $v(t) = 0$ and solve for $t$: $3t^2 - 12 = 0$ $t^2 = 4$ $t = 2 \quad (\text{since} \, t \geq 0)$

Now, substitute $t = 2$ into the acceleration function $a(t) = 6t$: $a(2) = 6(2) = 12 \, \text{m/s}^2$

Let me know if you want more details or have any questions!

Here are some follow-up questions:

1. What does the sign of acceleration tell you about the motion of the particle?
2. How would you find the time when the particle is at rest?
3. Can you derive the jerk (the rate of change of acceleration) from the acceleration function?
4. What is the significance of the particle's velocity being zero?
5. How does this equation of motion relate to a real-world physical scenario?

Tip: When calculating derivatives, remember that they represent rates of change; velocity is the rate of change of displacement, and acceleration is the rate of change of velocity.