To determine when the acceleration is zero, we first need to find the acceleration function by following these steps:

1. Position function:
   $s(t) = t^4 - 4t^3 - 20t^2 + 20t$

2. Velocity function:
   The velocity is the first derivative of the position function $s(t)$, i.e., $v(t) = s'(t)$: $v(t) = \frac{d}{dt}(t^4 - 4t^3 - 20t^2 + 20t) = 4t^3 - 12t^2 - 40t + 20$

3. Acceleration function:
   The acceleration is the derivative of the velocity function $v(t)$, i.e., $a(t) = v'(t)$: $a(t) = \frac{d}{dt}(4t^3 - 12t^2 - 40t + 20) = 12t^2 - 24t - 40$

4. Set the acceleration equal to zero: $12t^2 - 24t - 40 = 0$ Simplifying by dividing the equation by 4: $3t^2 - 6t - 10 = 0$

5. Solve the quadratic equation: Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 3$, $b = -6$, and $c = -10$: $t = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(-10)}}{2(3)} = \frac{6 \pm \sqrt{36 + 120}}{6} = \frac{6 \pm \sqrt{156}}{6} = \frac{6 \pm 2\sqrt{39}}{6}$ Simplifying further: $t = 1 \pm \frac{\sqrt{39}}{3}$

Since $t \geq 0$, the valid solutions are:

$t = 1 + \frac{\sqrt{39}}{3} \quad \text{or} \quad t = 1 - \frac{\sqrt{39}}{3}$

We discard the negative result because it would lead to negative time, which is not allowed in this context. Therefore, the acceleration is zero at $t = 1 + \frac{\sqrt{39}}{3}$ seconds.

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Would you like further clarification or more details on this problem? Here are 5 related questions for practice:

1. What is the velocity of the particle at $t = 0$?
2. How do you interpret the physical meaning of zero acceleration?
3. At what times is the velocity of the particle equal to zero?
4. What is the maximum velocity of the particle?
5. How does the concavity of the position function relate to acceleration?

Tip: Remember, acceleration is the rate of change of velocity. A zero acceleration means the velocity is constant at that moment.