The coordinates of a moving particle at any time t are given by, `x = 2t^(3)` and `y = 3t^(3)`. Acceleration of the particle is given by

Options

A. `468 t`
B. `t sqrt(468)`
C. `234 t^(2)`
D. `t sqrt(234)`

Explanation

To find the acceleration of the particle whose coordinates are given by \( x = 2t^3 \) and \( y = 3t^3 \), we can follow these steps:

### Step 1: Find the velocity components
The velocity components in the x and y directions can be found by taking the derivatives of the position functions with respect to time \( t \).

1. **Velocity in the x-direction**:
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(2t^3) = 6t^2 \]

2. **Velocity in the y-direction**:
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(3t^3) = 9t^2 \]

### Step 2: Find the acceleration components
The acceleration components can be found by taking the derivatives of the velocity components with respect to time \( t \).

1. **Acceleration in the x-direction**:
\[ a_x = \frac{dv_x}{dt} = \frac{d}{dt}(6t^2) = 12t \]

2. **Acceleration in the y-direction**:
\[ a_y = \frac{dv_y}{dt} = \frac{d}{dt}(9t^2) = 18t \]

### Step 3: Write the acceleration vector
The acceleration vector \( \vec{A} \) can be expressed in terms of its components:
\[ \vec{A} = a_x \hat{i} + a_y \hat{j} = (12t) \hat{i} + (18t) \hat{j} \]

### Step 4: Find the magnitude of the acceleration
The magnitude of the acceleration vector can be calculated using the Pythagorean theorem:
\[ |\vec{A}| = \sqrt{(a_x)^2 + (a_y)^2} = \sqrt{(12t)^2 + (18t)^2} \]
Calculating this gives:
\[ |\vec{A}| = \sqrt{144t^2 + 324t^2} = \sqrt{468t^2} = \sqrt{468} \cdot t \]

### Final Answer
Thus, the acceleration of the particle is:
\[ \vec{A} = (12t) \hat{i} + (18t) \hat{j} \quad \text{and} \quad |\vec{A}| = t\sqrt{468} \]

---