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.



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

### Given Information:
- The particle moves along the curve \( y = \frac{2x^{3/2}}{3} \).
- The speed of the particle is constant at 9 cm/sec in the positive \( x \)-direction.
- The position starts at the origin, meaning at \( t = 0 \), \( x = 0 \) and \( y = 0 \).

### Step 1: Speed along the x-axis
Since the particle moves at a speed of 9 cm/sec in the positive \( x \)-direction, we know the following:

\[
\frac{dx}{dt} = 9
\]

### Step 2: Solving for \( x(t) \)
Integrating this equation with respect to time \( t \), we get:

\[
x(t) = 9t + C
\]

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

\[
x(t) = 9t
\]

### Step 3: Finding \( y(t) \)
Now that we know \( x(t) = 9t \), we can substitute this into the curve equation to find \( y(t) \):

\[
y = \frac{2x^{3/2}}{3}
\]

Substituting \( x = 9t \):

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

Simplify the expression:

\[
y(t) = \frac{2(9^{3/2} \cdot t^{3/2})}{3}
\]

Since \( 9^{3/2} = 27 \), we have:

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

### Step 4: Position at \( t = 18 \)
Now, let's find \( x \) and \( y \) at \( t = 18 \).

For \( x(t) \):

\[
x(18) = 9 \times 18 = 162 \, \text{cm}
\]

For \( y(t) \):

\[
y(18) = 18 \times 18^{3/2}
\]

Since \( 18^{3/2} = 18 \times \sqrt{18} = 18 \times 4.242 \approx 76.356 \), we get:

\[
y(18) \approx 18 \times 76.356 = 1374.41 \, \text{cm}
\]

### Final Position:
At \( t = 18 \), the particle's position is approximately:

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

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 \( x \)- and \( y \)-components or if motion is constrained to one direction.

A particle starts at 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.

Determine the position of a particle moving along the curve y=(2x^3/2)/3 at a speed of 9 cm/sec in the positive x-direction. Learn how to calculate x(t) and y(t) at t=18 seconds using integration and parametric equations. Suitable for students in Grades 11-12.

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