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 given by:

Options

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

Answer

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 coordinates with respect to time
We start by differentiating the coordinates \( x \) and \( y \) with respect to time \( t \).

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

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

### Step 2: Write the velocity vector
The velocity vector \( \vec{v} \) can be expressed in terms of its components:
\[ \vec{v} = \frac{dx}{dt} \hat{i} + \frac{dy}{dt} \hat{j} \]
Substituting the derivatives we found:
\[ \vec{v} = (2ct) \hat{i} + (2bt) \hat{j} \]

### Step 3: Factor out common terms
We can factor out \( 2t \) from the velocity vector:
\[ \vec{v} = 2t (c \hat{i} + b \hat{j}) \]

### Step 4: Find the magnitude of the velocity vector
The speed of the particle is the magnitude of the velocity vector:
\[ |\vec{v}| = \sqrt{(2ct)^2 + (2bt)^2} \]
Calculating this:
\[ |\vec{v}| = \sqrt{4c^2t^2 + 4b^2t^2} = \sqrt{4t^2(c^2 + b^2)} = 2t \sqrt{c^2 + b^2} \]

### Step 5: Final expression for speed
Thus, the speed of the particle is given by:
\[ \text{Speed} = 2t \sqrt{c^2 + b^2} \]