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?

We are given the acceleration vector \(\mathbf{a}(t) = (0, 0, 6)\), the initial velocity \(\mathbf{v}(0) = (1, 1, -2)\), and the initial position \(\mathbf{c}(0) = (3, 4, 0)\). We need to find the velocity vector \(\mathbf{v}(t)\), the position vector \(\mathbf{c}(t)\), and determine the time(s) when the particle touches the xy-plane.

### Step 1: Find the velocity vector \(\mathbf{v}(t)\)
Acceleration is the derivative of velocity, i.e.,

\[
\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt}
\]

Since \(\mathbf{a}(t) = (0, 0, 6)\), integrate it with respect to time \(t\) to find the velocity vector \(\mathbf{v}(t)\):

\[
\mathbf{v}(t) = \int \mathbf{a}(t) \, dt = \int (0, 0, 6) \, dt = (0, 0, 6t) + \mathbf{v}_0
\]

where \(\mathbf{v}_0 = (1, 1, -2)\) is the initial velocity. So,

\[
\mathbf{v}(t) = (1, 1, -2) + (0, 0, 6t) = (1, 1, -2 + 6t)
\]

Thus, the velocity vector is:

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

### Step 2: Find the position vector \(\mathbf{c}(t)\)
Velocity is the derivative of position, i.e.,

\[
\mathbf{v}(t) = \frac{d\mathbf{c}(t)}{dt}
\]

Integrating the velocity vector \(\mathbf{v}(t) = (1, 1, -2 + 6t)\) with respect to time \(t\), we get the position vector \(\mathbf{c}(t)\):

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

This yields:

\[
\mathbf{c}(t) = (t, t, -2t + 3t^2) + \mathbf{c}_0
\]

where \(\mathbf{c}_0 = (3, 4, 0)\) is the initial position. So,

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

Thus, the position vector is:

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

### Step 3: Find when the particle touches the xy-plane
The particle touches the xy-plane when its \(z\)-coordinate is zero. From the position vector \(\mathbf{c}(t) = (t + 3, t + 4, -2t + 3t^2)\), the \(z\)-coordinate is:

\[
z(t) = -2t + 3t^2
\]

Set \(z(t) = 0\) to find when the particle touches the xy-plane:

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

Factor the equation:

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

Thus, \(t = 0\) or \(t = \frac{2}{3}\).

### Final Answer:

- The velocity vector is \(\mathbf{v}(t) = (1, 1, -2 + 6t)\).
- The position vector is \(\mathbf{c}(t) = (t + 3, t + 4, -2t + 3t^2)\).
- The particle touches the xy-plane at \(t = 0\) and \(t = \frac{2}{3}\).

Would you like more details or have any questions about the steps? Here are some related questions to expand the topic:

1. How do we derive the position vector from the velocity vector using integration?
2. What is the physical interpretation of acceleration in this problem?
3. How would the problem change if the initial velocity were different?
4. What does the time \(t = \frac{2}{3}\) represent in terms of motion and position?
5. How could the method be applied to higher-dimensional motion problems?

**Tip**: When integrating acceleration to find velocity or velocity to find position, always include the constant of integration, which can be determined using initial conditions.

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?

Learn how to find the velocity and position vectors for a particle with constant acceleration ~a(t) = (0, 0, 6) and initial conditions (1, 1, -2) for velocity and (3, 4, 0) for position. This problem also explores the time(s) when the particle touches the xy-plane, suitable for high school students in Grades 11-12.

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