The problem asks for the acceleration of a particle when the velocity of the particle is zero. The position of the particle is given as:

$x(t) = t^4 - 4t^3$

We will first find the velocity $v(t)$ by taking the first derivative of $x(t)$ with respect to $t$, and then the acceleration $a(t)$ by taking the second derivative of $x(t)$. Finally, we will find the time $t$ when the velocity is zero and use it to calculate the acceleration.

Let's solve this step-by-step.

Step 1: Find velocity $v(t)$

[ v(t) = \frac{d}{dt}[x(t)] = \frac{d}{dt}[t^4 - 4t^3] ]

$v(t) = 4t^3 - 12t^2$

Step 2: Set $v(t) = 0$ to find the time(s) when velocity is zero

$4t^3 - 12t^2 = 0$

Factor the equation: $4t^2(t - 3) = 0$

This gives: $t = 0 \quad \text{or} \quad t = 3$

Since the problem specifies $t > 0$, we take $t = 3$.

Step 3: Find acceleration $a(t)$

Acceleration is the derivative of velocity: [ a(t) = \frac{d}{dt}[v(t)] = \frac{d}{dt}[4t^3 - 12t^2] ]

$a(t) = 12t^2 - 24t$

Step 4: Substitute $t = 3$ into $a(t)$

$a(3) = 12(3)^2 - 24(3)$ $a(3) = 12(9) - 72$ $a(3) = 108 - 72 = 36$

Thus, the acceleration of the particle when the velocity is zero is $36$. The correct answer is (C) 36.

Would you like more details on any step? Here are some related questions to expand your understanding:

1. How do you interpret position, velocity, and acceleration functions in calculus?
2. What is the physical significance of finding when the velocity is zero?
3. How does factoring help in solving polynomial equations like $4t^2(t - 3) = 0$?
4. What is the role of derivatives in physics for describing motion?
5. Can velocity or acceleration be negative, and what would that mean physically?

Tip: When solving motion problems, always carefully check the conditions, such as whether time can be negative or zero in a specific context.