The coordinates of a moving particle at time t are given by `x=ct^(2)` and `y=bt^(2)`. The speed of the particle is given by :-

Options

A. `2t(c+b)`
B. `2tsqrt(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 can follow these steps:

### Step 1: Determine the expressions for 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 (\( v_x \))**:
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(ct^2) = 2ct \]

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

### Step 2: Calculate the speed of the particle
The speed \( v \) of the particle is given by the magnitude of the velocity vector, which can be calculated using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} \]

Substituting the expressions for \( v_x \) and \( v_y \):
\[ v = \sqrt{(2ct)^2 + (2bt)^2} \]

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

### Step 4: Factor out the square root
Taking the square root of the factors:
\[ v = \sqrt{4} \cdot \sqrt{t^2} \cdot \sqrt{c^2 + b^2} \]
\[ = 2t\sqrt{c^2 + b^2} \]

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