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 need to find the speed of the particle when the x-component of the acceleration is 36 m/s². We will follow these steps:

### Step 1: Find the expressions for velocity in the x and y directions.

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 functions with respect to time \( t \).

**Velocity in the x-direction (\( V_x \)):**
\[ V_x = \frac{dx}{dt} = \frac{d}{dt}(7 + 3t^3) = 0 + 9t^2 = 9t^2 \, \text{m/s} \]

**Velocity in the y-direction (\( V_y \)):**
\[ V_y = \frac{dy}{dt} = \frac{d}{dt}(13 + 5t - 9t^2) = 0 + 5 - 18t = 5 - 18t \, \text{m/s} \]

### Step 2: Find the expressions for acceleration in the x and y directions.

Next, we differentiate the velocity functions to find the acceleration components.

**Acceleration in the x-direction (\( A_x \)):**
\[ A_x = \frac{dV_x}{dt} = \frac{d}{dt}(9t^2) = 18t \, \text{m/s}^2 \]

**Acceleration in the y-direction (\( A_y \)):**
\[ A_y = \frac{dV_y}{dt} = \frac{d}{dt}(5 - 18t) = -18 \, \text{m/s}^2 \]

### Step 3: Set the x-component of acceleration equal to 36 m/s² and solve for time \( t \).

We know:
\[ A_x = 18t = 36 \, \text{m/s}^2 \]
Solving for \( t \):
\[ t = \frac{36}{18} = 2 \, \text{s} \]

### Step 4: Calculate the velocity components at \( t = 2 \) s.

Now, we substitute \( t = 2 \) s into the velocity equations to find \( V_x \) and \( V_y \).

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

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

### Step 5: Calculate the magnitude of the velocity (speed).

The speed is the magnitude of the velocity vector, calculated using the formula:
\[ \text{Speed} = \sqrt{V_x^2 + V_y^2} \]
Substituting the values:
\[ \text{Speed} = \sqrt{(36)^2 + (-31)^2} = \sqrt{1296 + 961} = \sqrt{2257} \]

### Final Result:
The speed of the particle when the x-component of the acceleration is \( 36 \, \text{m/s}^2 \) is:
\[ \text{Speed} = \sqrt{2257} \, \text{m/s} \]