To find the position function $s(t)$ of the particle given the acceleration function $a(t) = 3 \cos(t) - 9 \sin(t)$, we need to integrate $a(t)$ twice and use the initial conditions $s(0) = -3$ and $v(0) = 12$.

Step 1: Find the Velocity Function $v(t)$

1. Start by integrating $a(t)$: $v(t) = \int (3 \cos(t) - 9 \sin(t)) \, dt$

2. Integrate each term:

   • The integral of $3 \cos(t)$ is $3 \sin(t)$.
   • The integral of $-9 \sin(t)$ is $9 \cos(t)$.

   So, $v(t) = 3 \sin(t) + 9 \cos(t) + C$

3. Use the initial condition $v(0) = 12$ to solve for $C$: $v(0) = 3 \sin(0) + 9 \cos(0) + C = 12$ $0 + 9 + C = 12$ $C = 3$

   Therefore, $v(t) = 3 \sin(t) + 9 \cos(t) + 3$

Step 2: Find the Position Function $s(t)$

1. Now integrate $v(t)$ to find $s(t)$: $s(t) = \int (3 \sin(t) + 9 \cos(t) + 3) \, dt$

2. Integrate each term:

   • The integral of $3 \sin(t)$ is $-3 \cos(t)$.
   • The integral of $9 \cos(t)$ is $9 \sin(t)$.
   • The integral of $3$ is $3t$.

   So, $s(t) = -3 \cos(t) + 9 \sin(t) + 3t + D$

3. Use the initial condition $s(0) = -3$ to solve for $D$: $s(0) = -3 \cos(0) + 9 \sin(0) + 3(0) + D = -3$ $-3 + 0 + 0 + D = -3$ $D = 0$

   Thus, the position function is: $s(t) = -3 \cos(t) + 9 \sin(t) + 3t$

Final Answer

The position function for the particle is: $s(t) = -3 \cos(t) + 9 \sin(t) + 3t$

Would you like further explanation on any of these steps?

Here are some related questions for deeper understanding:

1. How does the initial condition affect the integration constants?
2. What would the solution look like if $s(0)$ and $v(0)$ were different values?
3. Can you solve this problem using alternative methods like the Laplace transform?
4. How does acceleration relate to the curvature of the position function?
5. What is the physical interpretation of each term in the position function?

Tip: Always double-check initial conditions after integration to ensure accuracy in finding constants.