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

Options

A. `2t sqrt(b^(2)-c^(2))`
B. `sqrt(4b^(2)t^(2)+c^(2))`
C. `2t(b+c)`
D. `2t(b-c)`

Explanation

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

### Step 1: Determine the velocity components
The velocity of the particle can be broken down into its components in the x and y directions.

1. **Horizontal component of velocity (\( v_x \))**:
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(ct) = c \]

2. **Vertical component of velocity (\( v_y \))**:
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(bt^2) = 2bt \]

### Step 2: Write the velocity vector
The velocity vector \( \vec{v} \) can be expressed in terms of its components:
\[ \vec{v} = v_x \hat{i} + v_y \hat{j} = c \hat{i} + 2bt \hat{j} \]

### Step 3: Calculate the magnitude of the velocity
The speed of the particle is the magnitude of the velocity vector, which can be calculated using the Pythagorean theorem:
\[ |\vec{v}| = \sqrt{v_x^2 + v_y^2} \]
Substituting the values of \( v_x \) and \( v_y \):
\[ |\vec{v}| = \sqrt{c^2 + (2bt)^2} = \sqrt{c^2 + 4b^2t^2} \]

### Final Answer
Thus, the speed of the particle is given by:
\[ \text{Speed} = \sqrt{c^2 + 4b^2t^2} \]

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