A particle is moving with speed `v=bsqrt(x)` along positive x-axis. Calculate the speed of the particle at time `t=tau` (assume that the particle is at origin at t = 0).

Options

A. `(b^(2)tau)/(sqrt(2))`
B. `(b^(2)tau)/4`
C. `(b^(2)tau)/2`
D. `b^(2)tau`

Explanation

To solve the problem, we need to find the speed of a particle moving with speed \( v = b\sqrt{x} \) along the positive x-axis, given that the particle is at the origin at \( t = 0 \).

### Step-by-Step Solution:

1. **Understand the relationship between speed and position:**
The speed \( v \) of the particle is given by:
\[ v = \frac{dx}{dt} = b\sqrt{x} \]

2. **Separate variables and integrate:**
Rearranging the equation gives:
\[ \frac{dx}{\sqrt{x}} = b \, dt \]
Now, we integrate both sides. The left side can be integrated as follows:
\[ \int \frac{dx}{\sqrt{x}} = 2\sqrt{x} \]
The right side integrates to:
\[ \int b \, dt = bt \]
Therefore, we have:
\[ 2\sqrt{x} = bt + C \]
Since the particle is at the origin when \( t = 0 \) (i.e., \( x = 0 \)), we find that \( C = 0 \). Thus:
\[ 2\sqrt{x} = bt \]

3. **Solve for \( x \):**
Squaring both sides gives:
\[ 4x = b^2t^2 \quad \Rightarrow \quad x = \frac{b^2t^2}{4} \]

4. **Find the speed at time \( t = \tau \):**
We need to find the speed \( v \) at \( t = \tau \). We already have the expression for speed:
\[ v = b\sqrt{x} \]
Substituting \( x = \frac{b^2\tau^2}{4} \):
\[ v = b\sqrt{\frac{b^2\tau^2}{4}} = b \cdot \frac{b\tau}{2} = \frac{b^2\tau}{2} \]

5. **Final Answer:**
The speed of the particle at time \( t = \tau \) is:
\[ v = \frac{b^2\tau}{2} \]