The coordinates of a particle moving in XY-plane at any instant of time t are `x=4t^(2),y=3t^(2)`. The speed of the particle at that instant is

Explanation

To find the speed of the particle at any instant of time \( t \), we will follow these steps:

### Step 1: Write down the equations of motion
The coordinates of the particle are given as:
\[ x = 4t^2 \]
\[ y = 3t^2 \]

### Step 2: Differentiate the position equations to find velocities
To find the velocity in the x-direction (\( v_x \)), we differentiate \( x \) with respect to \( t \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(4t^2) = 8t \]

Next, we find the velocity in the y-direction (\( v_y \)) by differentiating \( y \) with respect to \( t \):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(3t^2) = 6t \]

### Step 3: Calculate the speed of the particle
The speed of the particle is given by the magnitude of the velocity vector, which can be calculated using the formula:
\[ \text{Speed} = \sqrt{v_x^2 + v_y^2} \]

Substituting the expressions for \( v_x \) and \( v_y \):
\[ \text{Speed} = \sqrt{(8t)^2 + (6t)^2} \]

Calculating the squares:
\[ = \sqrt{64t^2 + 36t^2} \]
\[ = \sqrt{100t^2} \]

Taking the square root:
\[ = 10t \]

### Final Answer
Thus, the speed of the particle at any instant of time \( t \) is:
\[ \text{Speed} = 10t \text{ meters per second} \]

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