The co-ordinates of a moving particles at time t, given by `x = at^(2), y = bt^(2)`. The speed of the particle is

Options

A. `2 (a + b)t`
B. `(a^(2) + b^(2))^(1//2)` t
C. `2(a^(2) + b^(2))^(1//2)t`
D. `(a + b)t`

Answer

To find the speed of the particle given its coordinates as \( x = at^2 \) and \( y = bt^2 \), we can follow these steps:

### Step 1: Write down the position vector
The position vector \( \vec{r} \) of the particle can be expressed in terms of its coordinates:
\[ \vec{r} = x \hat{i} + y \hat{j} = (at^2) \hat{i} + (bt^2) \hat{j} \]

### Step 2: Differentiate the position vector to find velocity
To find the velocity \( \vec{v} \), we differentiate the position vector \( \vec{r} \) with respect to time \( t \):
\[ \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}[(at^2) \hat{i} + (bt^2) \hat{j}] \]
Using the power rule of differentiation:
\[ \vec{v} = 2at \hat{i} + 2bt \hat{j} \]

### Step 3: Calculate the magnitude of the velocity vector
The speed of the particle is the magnitude of the velocity vector \( \vec{v} \):
\[ |\vec{v}| = \sqrt{(2at)^2 + (2bt)^2} \]
This simplifies to:
\[ |\vec{v}| = \sqrt{4a^2t^2 + 4b^2t^2} = \sqrt{4t^2(a^2 + b^2)} = 2t\sqrt{a^2 + b^2} \]

### Step 4: Write the final expression for speed
Thus, the speed of the particle is given by:
\[ \text{Speed} = 2t\sqrt{a^2 + b^2} \]

### Conclusion
The speed of the particle is \( 2t\sqrt{a^2 + b^2} \).