Given that the acceleration vector is a(t)=(-9cos(3t))I+(-9sin(3t))J+(3t)K
, the initial velocity is v(0)=I+K
, and the initial position vector is X(0)=I+J+K
, compute: The velocity vector and position vector

To compute the velocity vector \(\vec{v}(t)\) and position vector \(\vec{X}(t)\), we need to integrate the acceleration vector \(\vec{a}(t)\) with respect to time \(t\), using the initial conditions provided.

### Given:
1. \(\vec{a}(t) = (-9\cos(3t)) \hat{i} + (-9\sin(3t)) \hat{j} + (3t) \hat{k}\)
2. Initial velocity \(\vec{v}(0) = \hat{i} + \hat{k}\)
3. Initial position \(\vec{X}(0) = \hat{i} + \hat{j} + \hat{k}\)

#### Step 1: Compute the velocity vector \(\vec{v}(t)\)

The velocity vector is obtained by integrating the acceleration vector:

\[
\vec{v}(t) = \int \vec{a}(t) \, dt
\]

We integrate each component of \(\vec{a}(t)\):

- \(v_x(t) = \int (-9\cos(3t)) \, dt = -3\sin(3t) + C_1\)
- \(v_y(t) = \int (-9\sin(3t)) \, dt = 3\cos(3t) + C_2\)
- \(v_z(t) = \int 3t \, dt = \frac{3}{2}t^2 + C_3\)

Now, we use the initial condition \(\vec{v}(0) = \hat{i} + \hat{k}\), which gives:

- \(v_x(0) = 1\), so \( -3\sin(0) + C_1 = 1 \) \(\Rightarrow C_1 = 1\)
- \(v_y(0) = 0\), so \( 3\cos(0) + C_2 = 0 \) \(\Rightarrow C_2 = -3\)
- \(v_z(0) = 1\), so \(\frac{3}{2}(0)^2 + C_3 = 1 \) \(\Rightarrow C_3 = 1\)

Thus, the velocity vector becomes:

\[
\vec{v}(t) = (-3\sin(3t) + 1) \hat{i} + (3\cos(3t) - 3) \hat{j} + \left(\frac{3}{2}t^2 + 1\right) \hat{k}
\]

#### Step 2: Compute the position vector \(\vec{X}(t)\)

The position vector is obtained by integrating the velocity vector:

\[
\vec{X}(t) = \int \vec{v}(t) \, dt
\]

We integrate each component of \(\vec{v}(t)\):

- \(x(t) = \int (-3\sin(3t) + 1) \, dt = \cos(3t) + t + D_1\)
- \(y(t) = \int (3\cos(3t) - 3) \, dt = \sin(3t) - 3t + D_2\)
- \(z(t) = \int \left(\frac{3}{2}t^2 + 1\right) \, dt = \frac{1}{2}t^3 + t + D_3\)

Now, we use the initial condition \(\vec{X}(0) = \hat{i} + \hat{j} + \hat{k}\), which gives:

- \(x(0) = 1\), so \(\cos(0) + 0 + D_1 = 1 \Rightarrow D_1 = 0\)
- \(y(0) = 1\), so \(\sin(0) - 0 + D_2 = 1 \Rightarrow D_2 = 1\)
- \(z(0) = 1\), so \(\frac{1}{2}(0)^3 + 0 + D_3 = 1 \Rightarrow D_3 = 1\)

Thus, the position vector becomes:

\[
\vec{X}(t) = (\cos(3t) + t) \hat{i} + (\sin(3t) - 3t + 1) \hat{j} + \left(\frac{1}{2}t^3 + t + 1\right) \hat{k}
\]

### Final Answer:

- **Velocity vector**:
\[
\vec{v}(t) = (-3\sin(3t) + 1) \hat{i} + (3\cos(3t) - 3) \hat{j} + \left(\frac{3}{2}t^2 + 1\right) \hat{k}
\]

- **Position vector**:
\[
\vec{X}(t) = (\cos(3t) + t) \hat{i} + (\sin(3t) - 3t + 1) \hat{j} + \left(\frac{1}{2}t^3 + t + 1\right) \hat{k}
\]

Would you like more details on any of these steps, or do you have any questions?

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**Related questions:**
1. How would the problem change if the initial velocity were different?
2. What is the physical interpretation of the cosine and sine functions in the velocity?
3. How would you find the time when the particle returns to its initial position?
4. What is the speed of the particle as a function of time?
5. How does the velocity change for large values of \(t\)?

**Tip:** To find the total distance traveled by the particle, you need to integrate the speed over time, which is the magnitude of the velocity vector.

Given that the acceleration vector is a(t)=(-9cos(3t))I+(-9sin(3t))J+(3t)K, the initial velocity is v(0)=I+K, and the initial position vector is X(0)=I+J+K, compute: The velocity vector and position vector

This solution demonstrates how to compute the velocity and position vectors given an acceleration vector in a 3D space. Using vector calculus, the problem involves integration of trigonometric and polynomial functions, along with initial conditions, making it suitable for undergraduate students in calculus-based physics.

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