The co-ordintes of a mving particle at any time t are given by
`x = ct^(2) and y = bt^(2)`
The Speedof the particle is given by

Options

A. `t (c + b)`
B. `2t sqrt((c^(2)-b^(2)))`
C. `tsqrt((c^(2) + b^(2)))`
D. `2tsqrt((c^(2) + b^(2)))`

Explanation

To find the speed of the particle whose coordinates are given by \( x = ct^2 \) and \( y = bt^2 \), we will follow these steps:

### Step 1: Differentiate the position functions
The first step is to find the velocities in the x and y directions by differentiating the position functions with respect to time \( t \).

1. Differentiate \( x = ct^2 \):
\[ \frac{dx}{dt} = 2ct \]

2. Differentiate \( y = bt^2 \):
\[ \frac{dy}{dt} = 2bt \]

### Step 2: Find the resultant speed
The speed of the particle is the magnitude of the velocity vector, which can be calculated using the Pythagorean theorem. The speed \( v \) is given by:
\[ v = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \]

Substituting the derivatives we found:
\[ v = \sqrt{(2ct)^2 + (2bt)^2} \]

### Step 3: Simplify the expression
Now, simplify the expression inside the square root:
\[ v = \sqrt{4c^2t^2 + 4b^2t^2} \]
\[ v = \sqrt{4t^2(c^2 + b^2)} \]

### Step 4: Factor out the constant
We can factor out the constant \( 4t^2 \) from the square root:
\[ v = \sqrt{4} \cdot \sqrt{t^2} \cdot \sqrt{c^2 + b^2} \]
\[ v = 2t \sqrt{c^2 + b^2} \]

### Final Answer
Thus, the speed of the particle is:
\[ v = 2t \sqrt{c^2 + b^2} \]