) Find the velocity and position vectors, ~v(t) and ~c(t), given that ~a(t) = (0, 0, 6) with an initial velocity of (1, 1, −2) and initial position (3, 4, 0). For what value(s) of t does the particle touch the xy-plane?

To find the velocity vector \(\mathbf{v}(t)\) and the position vector \(\mathbf{c}(t)\), we start with the given acceleration vector \(\mathbf{a}(t) = (0, 0, 6)\).

### Step 1: Find the velocity vector \(\mathbf{v}(t)\)

The velocity vector is found by integrating the acceleration vector with respect to time \(t\):

\[
\mathbf{v}(t) = \int \mathbf{a}(t) \, dt = \int (0, 0, 6) \, dt = (0, 0, 6t) + \mathbf{C}
\]

where \(\mathbf{C}\) is the constant of integration. We know the initial velocity:

\[
\mathbf{v}(0) = (1, 1, -2)
\]

Substituting \(t = 0\):

\[
(0, 0, 0) + \mathbf{C} = (1, 1, -2) \implies \mathbf{C} = (1, 1, -2)
\]

Thus, the velocity vector is:

\[
\mathbf{v}(t) = (1, 1, 6t - 2)
\]

### Step 2: Find the position vector \(\mathbf{c}(t)\)

Next, we integrate the velocity vector to find the position vector:

\[
\mathbf{c}(t) = \int \mathbf{v}(t) \, dt = \int (1, 1, 6t - 2) \, dt = (t, t, 3t^2 - 2t) + \mathbf{D}
\]

where \(\mathbf{D}\) is another constant of integration. We know the initial position:

\[
\mathbf{c}(0) = (3, 4, 0)
\]

Substituting \(t = 0\):

\[
(0, 0, 0) + \mathbf{D} = (3, 4, 0) \implies \mathbf{D} = (3, 4, 0)
\]

Thus, the position vector is:

\[
\mathbf{c}(t) = (t + 3, t + 4, 3t^2 - 2t)
\]

### Step 3: Determine when the particle touches the xy-plane

The particle touches the xy-plane when the \(z\)-coordinate of the position vector \(\mathbf{c}(t)\) is zero:

\[
3t^2 - 2t = 0
\]

Factoring out \(t\):

\[
t(3t - 2) = 0
\]

This gives us two solutions:

1. \(t = 0\)
2. \(3t - 2 = 0 \implies t = \frac{2}{3}\)

### Summary

The velocity vector is:

\[
\mathbf{v}(t) = (1, 1, 6t - 2)
\]

The position vector is:

\[
\mathbf{c}(t) = (t + 3, t + 4, 3t^2 - 2t)
\]

The particle touches the xy-plane at:

- \(t = 0\) (initial position)
- \(t = \frac{2}{3}\)

If you have any questions or need further details, feel free to ask! Here are some related questions for further exploration:

1. What is the acceleration vector when \(t = \frac{2}{3}\)?
2. How would the motion change if the initial velocity were different?
3. What is the path traced by the particle in the xy-plane?
4. How would the position vector change with a different acceleration?
5. Can you find the time when the particle reaches a certain height? 

**Tip:** Always check your integration constants with initial conditions to ensure accuracy.

Find the velocity and position vectors, ~v(t) and ~c(t), given that ~a(t) = (0, 0, 6) with an initial velocity of (1, 1, −2) and initial position (3, 4, 0). For what value(s) of t does the particle touch the xy-plane?

Explore how to calculate the velocity and position vectors from a given acceleration vector in this step-by-step solution. This problem involves integrating the acceleration vector, applying initial conditions, and determining when the particle touches the xy-plane. Suitable for high school students in Grades 11-12.

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