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 and the position vector given the acceleration vector , we will integrate the acceleration function step by step.
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Finding the Velocity Vector: The velocity is the integral of the acceleration :
Where is the constant of integration. To find , we use the initial condition for velocity :
Therefore, the velocity vector is:
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Finding the Position Vector: The position is the integral of the velocity :
Where is another constant of integration. Using the initial position :
Thus, the position vector is:
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Finding When the Particle Touches the xy-plane: The particle touches the xy-plane when the z-component of the position vector is zero:
Factoring gives:
Thus, the solutions are:
Summary:
- The velocity vector is:
- The position vector is:
- The particle touches the xy-plane at and .
Do you want further details on any part of this solution? Here are some related questions you might consider:
- How do you derive the equations of motion from acceleration?
- What do the terms in the velocity vector represent physically?
- How would the results change with a different initial velocity?
- Can you provide a graphical representation of the motion?
- 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.
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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
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