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 given the coordinates \( x = 2t^3 \) and \( y = 3t^3 \), we will follow these steps:

### Step 1: Find the velocity components
The velocity in the x-direction (\( v_x \)) is the derivative of \( x \) with respect to time \( t \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(2t^3) = 6t^2 \]

The velocity in the y-direction (\( v_y \)) is the derivative of \( y \) with respect to time \( t \):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(3t^3) = 9t^2 \]

### Step 2: Find the acceleration components
The acceleration in the x-direction (\( a_x \)) is the derivative of \( v_x \) with respect to time \( t \):
\[ a_x = \frac{dv_x}{dt} = \frac{d}{dt}(6t^2) = 12t \]

The acceleration in the y-direction (\( a_y \)) is the derivative of \( v_y \) with respect to time \( t \):
\[ a_y = \frac{dv_y}{dt} = \frac{d}{dt}(9t^2) = 18t \]

### Step 3: Calculate the magnitude of acceleration
The magnitude of the acceleration \( a \) can be found using the Pythagorean theorem:
\[ a = \sqrt{a_x^2 + a_y^2} \]
Substituting the values of \( a_x \) and \( a_y \):
\[ a = \sqrt{(12t)^2 + (18t)^2} = \sqrt{144t^2 + 324t^2} = \sqrt{468t^2} \]
\[ a = t \sqrt{468} \]

### Final Answer
Thus, the acceleration of the particle is:
\[ a = t \sqrt{468} \]

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