To solve this question, we need to calculate the acceleration of the particle at $t = 4$.

The position of the particle is given by $s(t) = t^2 + 4t + 4$. Acceleration is the second derivative of the position function $s(t)$ with respect to time $t$.

Steps:

1. Find the velocity function $v(t)$: The velocity is the first derivative of $s(t)$: $v(t) = \frac{ds(t)}{dt} = 2t + 4$

2. Find the acceleration function $a(t)$: The acceleration is the derivative of $v(t)$: $a(t) = \frac{dv(t)}{dt} = 2$

3. Evaluate acceleration at $t = 4$: Since the acceleration function is constant, $a(t) = 2$ for all $t$.

Final Answer:

The acceleration of the particle at $t = 4$ is $2$ (Option B).

Would you like further details about derivatives or related motion concepts? Here are five related questions for deeper understanding:

1. What is the significance of acceleration being constant in this problem?
2. How do position, velocity, and acceleration functions relate in general motion analysis?
3. What would the velocity be at $t = 4$ for this problem?
4. Can a particle's velocity and acceleration have opposite signs? What does it mean physically?
5. How can we graphically interpret $s(t)$, $v(t)$, and $a(t)$ for better understanding?

Tip:

To verify your derivative, always double-check your steps using the power rule: $\frac{d}{dt}[t^n] = n \cdot t^{n-1}$.