Math Problem Statement

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?

Solution

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

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

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

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

    v(0)=(0,0,0)+C=(1,1,2)    C=(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:

    v(t)=(1,1,6t2)\mathbf{v}(t) = (1, 1, 6t - 2)

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

    c(t)=v(t)dt=(1,1,6t2)dt=(t,t,3t22t)+D\mathbf{c}(t) = \int \mathbf{v}(t) \, dt = \int (1, 1, 6t - 2) \, dt = (t, t, 3t^2 - 2t) + \mathbf{D}

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

    c(0)=(0,0,0)+D=(3,4,0)    D=(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:

    c(t)=(t+3,t+4,3t22t)\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:

    3t22t=03t^2 - 2t = 0

    Factoring gives:

    t(3t2)=0t(3t - 2) = 0

    Thus, the solutions are:

    t=0or3t2=0    t=23t = 0 \quad \text{or} \quad 3t - 2 = 0 \implies t = \frac{2}{3}

Summary:

  • The velocity vector is: v(t)=(1,1,6t2)\mathbf{v}(t) = (1, 1, 6t - 2)
  • The position vector is: c(t)=(t+3,t+4,3t22t)\mathbf{c}(t) = (t + 3, t + 4, 3t^2 - 2t)
  • The particle touches the xy-plane at t=0t = 0 and t=23t = \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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Vector Calculus
Kinematics

Formulas

v(t) = ∫a(t) dt
c(t) = ∫v(t) dt

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12