Math Problem Statement
a particle starts the origin and moves along the curve y=(2x^3/2)/3 in the positive x-direction at a speed of 9 cm/sec, where x and y are in cm. Find the position of the particle at t=18.
Solution
To find the position of the particle at seconds, we need to determine the particle's position in terms of and .
Given Information:
- The particle moves along the curve .
- The speed of the particle is constant at 9 cm/sec in the positive -direction.
- The position starts at the origin, meaning at , and .
Step 1: Speed along the x-axis
Since the particle moves at a speed of 9 cm/sec in the positive -direction, we know the following:
Step 2: Solving for
Integrating this equation with respect to time , we get:
Since the particle starts at the origin, when , , so . Therefore, the position along the -axis as a function of time is:
Step 3: Finding
Now that we know , we can substitute this into the curve equation to find :
Substituting :
Simplify the expression:
Since , we have:
Step 4: Position at
Now, let's find and at .
For :
For :
Since , we get:
Final Position:
At , the particle's position is approximately:
Let me know if you want any more details or clarifications!
5 Relative Questions:
- How would the position change if the speed were different?
- What is the particle's trajectory in the y-direction over time?
- Can you derive a general formula for the velocity vector?
- What happens to the path of the particle if the curve equation changes?
- How would you determine the total distance traveled by the particle?
Tip:
When dealing with particle motion, always consider whether the velocity vector has both - and -components or if motion is constrained to one direction.
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Math Problem Analysis
Mathematical Concepts
Calculus
Differentiation
Integration
Parametric Equations
Formulas
Speed formula: dx/dt = speed
Position formula: x(t) = 9t
Curve equation: y = (2x^(3/2))/3
y(t) calculation: y(t) = 18 * t^(3/2)
Theorems
Fundamental Theorem of Calculus
Suitable Grade Level
Grades 11-12
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