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

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Identify the given equations**:
The coordinates of the particle are given as:
\[ x = ct^2 \]
\[ y = bt^2 \]

2. **Differentiate the position with respect to time**:
To find the velocity components, we need to differentiate \( x \) and \( y \) with respect to time \( t \).

- For the x-component:
\[ \frac{dx}{dt} = \frac{d}{dt}(ct^2) = c \cdot \frac{d}{dt}(t^2) = c \cdot 2t = 2ct \]

- For the y-component:
\[ \frac{dy}{dt} = \frac{d}{dt}(bt^2) = b \cdot \frac{d}{dt}(t^2) = b \cdot 2t = 2bt \]

3. **Calculate the speed**:
The speed of the particle is the magnitude of the velocity vector, which can be calculated using the Pythagorean theorem:
\[ \text{Speed} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \]
Substituting the components we found:
\[ \text{Speed} = \sqrt{(2ct)^2 + (2bt)^2} \]

4. **Simplify the expression**:
\[ \text{Speed} = \sqrt{4c^2t^2 + 4b^2t^2} \]
Factor out the common term:
\[ \text{Speed} = \sqrt{4t^2(c^2 + b^2)} = 2t\sqrt{c^2 + b^2} \]

5. **Final result**:
The speed of the particle is:
\[ \text{Speed} = 2t\sqrt{c^2 + b^2} \]