The position of a particle is given by `x=7+ 3t^(3)` m and `y=13+5t-9t^(2)m`, where x and y are the position coordinates, and t is the time in s. Find the speed (magnitude of the velocity) when the x component of the acceleration is `36 m//s^(2)`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Find the velocity components
The position of the particle is given by:
- \( x = 7 + 3t^3 \)
- \( y = 13 + 5t - 9t^2 \)

To find the velocity components, we differentiate the position equations with respect to time \( t \).

**Velocity in the x-direction (\( v_x \))**:
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(7 + 3t^3) = 9t^2 \]

**Velocity in the y-direction (\( v_y \))**:
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(13 + 5t - 9t^2) = 5 - 18t \]

### Step 2: Find the acceleration component
Next, we differentiate the velocity components to find the acceleration components.

**Acceleration in the x-direction (\( a_x \))**:
\[ a_x = \frac{dv_x}{dt} = \frac{d}{dt}(9t^2) = 18t \]

### Step 3: Set the x-component of acceleration to 36 m/s²
We need to find the time \( t \) when the x-component of acceleration \( a_x = 36 \, \text{m/s}^2 \).

Setting the equation:
\[ 18t = 36 \]

Solving for \( t \):
\[ t = \frac{36}{18} = 2 \, \text{s} \]

### Step 4: Calculate the velocity components at \( t = 2 \, \text{s} \)
Now we will find the velocity components at \( t = 2 \, \text{s} \).

**Calculate \( v_x \)**:
\[ v_x = 9t^2 = 9(2^2) = 9 \times 4 = 36 \, \text{m/s} \]

**Calculate \( v_y \)**:
\[ v_y = 5 - 18t = 5 - 18(2) = 5 - 36 = -31 \, \text{m/s} \]

### Step 5: Find the magnitude of the velocity
The magnitude of the velocity \( v \) can be calculated using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ v = \sqrt{(36)^2 + (-31)^2} = \sqrt{1296 + 961} = \sqrt{2257} \]
Calculating the square root:
\[ v \approx 47.51 \, \text{m/s} \]

### Final Answer
The speed (magnitude of the velocity) when the x-component of the acceleration is \( 36 \, \text{m/s}^2 \) is approximately \( 47.51 \, \text{m/s} \).

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