A body moves in a plane so that the displacements along the x and y axes are given by `x = 3t^3 and y = 4t^3`. The velocity of the body is

Options

A. 9t
B. 15t
C. `15t^2`
D. `25t^2`

Explanation

To find the velocity of the body moving in a plane with displacements given by \( x = 3t^3 \) and \( y = 4t^3 \), we will follow these steps:

### Step 1: Differentiate the displacement equations to find the velocities.
The velocity in the x-direction (\( v_x \)) is the derivative of the displacement in the x-direction with respect to time (\( t \)):
\[ v_x = \frac{dx}{dt} = \frac{d(3t^3)}{dt} = 9t^2 \]

The velocity in the y-direction (\( v_y \)) is the derivative of the displacement in the y-direction with respect to time (\( t \)):
\[ v_y = \frac{dy}{dt} = \frac{d(4t^3)}{dt} = 12t^2 \]

### Step 2: Calculate the resultant velocity.
The resultant velocity (\( v \)) can be found using the Pythagorean theorem since the x and y components are perpendicular:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values of \( v_x \) and \( v_y \):
\[ v = \sqrt{(9t^2)^2 + (12t^2)^2} \]
\[ v = \sqrt{81t^4 + 144t^4} \]
\[ v = \sqrt{225t^4} \]
\[ v = 15t^2 \]

### Step 3: State the final answer.
The velocity of the body is:
\[ v = 15t^2 \text{ m/s} \]

---