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 t=18t = 18 seconds, we need to determine the particle's position in terms of x(t)x(t) and y(t)y(t).

Given Information:

  • The particle moves along the curve y=2x3/23y = \frac{2x^{3/2}}{3}.
  • The speed of the particle is constant at 9 cm/sec in the positive xx-direction.
  • The position starts at the origin, meaning at t=0t = 0, x=0x = 0 and y=0y = 0.

Step 1: Speed along the x-axis

Since the particle moves at a speed of 9 cm/sec in the positive xx-direction, we know the following:

dxdt=9\frac{dx}{dt} = 9

Step 2: Solving for x(t)x(t)

Integrating this equation with respect to time tt, we get:

x(t)=9t+Cx(t) = 9t + C

Since the particle starts at the origin, when t=0t = 0, x=0x = 0, so C=0C = 0. Therefore, the position along the xx-axis as a function of time is:

x(t)=9tx(t) = 9t

Step 3: Finding y(t)y(t)

Now that we know x(t)=9tx(t) = 9t, we can substitute this into the curve equation to find y(t)y(t):

y=2x3/23y = \frac{2x^{3/2}}{3}

Substituting x=9tx = 9t:

y(t)=2(9t)3/23y(t) = \frac{2(9t)^{3/2}}{3}

Simplify the expression:

y(t)=2(93/2t3/2)3y(t) = \frac{2(9^{3/2} \cdot t^{3/2})}{3}

Since 93/2=279^{3/2} = 27, we have:

y(t)=2(27t3/2)3=18t3/2y(t) = \frac{2(27t^{3/2})}{3} = 18t^{3/2}

Step 4: Position at t=18t = 18

Now, let's find xx and yy at t=18t = 18.

For x(t)x(t):

x(18)=9×18=162cmx(18) = 9 \times 18 = 162 \, \text{cm}

For y(t)y(t):

y(18)=18×183/2y(18) = 18 \times 18^{3/2}

Since 183/2=18×18=18×4.24276.35618^{3/2} = 18 \times \sqrt{18} = 18 \times 4.242 \approx 76.356, we get:

y(18)18×76.356=1374.41cmy(18) \approx 18 \times 76.356 = 1374.41 \, \text{cm}

Final Position:

At t=18t = 18, the particle's position is approximately:

(x,y)=(162,1374.41)cm(x, y) = (162, 1374.41) \, \text{cm}

Let me know if you want any more details or clarifications!

5 Relative Questions:

  1. How would the position change if the speed were different?
  2. What is the particle's trajectory in the y-direction over time?
  3. Can you derive a general formula for the velocity vector?
  4. What happens to the path of the particle if the curve equation changes?
  5. 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 xx- and yy-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