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 \(\mathbf{c}(t)\) given the acceleration vector \(\mathbf{a}(t) = (0, 0, 6)\), we will integrate the acceleration function step by step.

1. **Finding the Velocity Vector**:
   The velocity \(\mathbf{v}(t)\) is the integral of the acceleration \(\mathbf{a}(t)\):

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

   Where \(\mathbf{C}\) is the constant of integration. To find \(\mathbf{C}\), we use the initial condition for velocity \(\mathbf{v}(0) = (1, 1, -2)\):

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

   Therefore, the velocity vector is:

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

2. **Finding the Position Vector**:
   The position \(\mathbf{c}(t)\) is the integral of the velocity \(\mathbf{v}(t)\):

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

   Where \(\mathbf{D}\) is another constant of integration. Using the initial position \(\mathbf{c}(0) = (3, 4, 0)\):

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

   Thus, the position vector is:

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

3. **Finding When the Particle Touches the xy-plane**:
   The particle touches the xy-plane when the z-component of the position vector is zero:

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

   Factoring gives:

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

   Thus, the solutions are:

   \[
   t = 0 \quad \text{or} \quad 3t - 2 = 0 \implies t = \frac{2}{3}
   \]

### 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 \(t = 0\) and \(t = \frac{2}{3}\).

Do you want further details on any part of this solution? Here are some related questions you might consider:

1. How do you derive the equations of motion from acceleration?
2. What do the terms in the velocity vector represent physically?
3. How would the results change with a different initial velocity?
4. Can you provide a graphical representation of the motion?
5. What are the implications of the particle's motion after it touches the xy-plane?

**Tip**: Always check initial conditions when integrating to find constants of integration.

Explore how to derive the velocity vector v(t) and position vector c(t) from the acceleration a(t) = (0, 0, 6) with initial conditions. Understand when the particle touches the xy-plane at t = 0 and t = 2/3 through detailed calculations. Suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation