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

Options

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

Answer

To solve the problem, we need to find the speed of a particle at time \( t = \tau \), given that its velocity \( v \) is defined as \( v = b \sqrt{x} \) and that it starts from the origin at \( t = 0 \).

### Step-by-Step Solution:

1. **Understanding the Given Velocity**:
The velocity of the particle is given by the equation:
\[ v = b \sqrt{x} \]

2. **Relating Velocity to Acceleration**:
We know that acceleration \( a \) can be expressed as:
\[ a = \frac{dv}{dt} \]
Using the chain rule, we can express this as:
\[ a = \frac{dv}{dx} \cdot \frac{dx}{dt} \]
Here, \( \frac{dx}{dt} \) is simply the velocity \( v \).

3. **Finding \( \frac{dv}{dx} \)**:
Differentiate the velocity \( v = b \sqrt{x} \) with respect to \( x \):
\[ \frac{dv}{dx} = b \cdot \frac{1}{2} x^{-1/2} = \frac{b}{2\sqrt{x}} \]

4. **Substituting into the Acceleration Formula**:
Now substitute \( \frac{dv}{dx} \) and \( v \) into the acceleration formula:
\[ a = \frac{b}{2\sqrt{x}} \cdot v \]
Since \( v = b \sqrt{x} \), we can substitute this in:
\[ a = \frac{b}{2\sqrt{x}} \cdot b \sqrt{x} = \frac{b^2}{2} \]
Thus, the acceleration \( a \) is constant and equal to \( \frac{b^2}{2} \).

5. **Using Kinematic Equation**:
We can use the kinematic equation to find the velocity at time \( t = \tau \):
\[ v = u + at \]
Since the initial velocity \( u = 0 \) (the particle starts from rest at the origin), we have:
\[ v(\tau) = 0 + \left(\frac{b^2}{2}\right) \tau = \frac{b^2}{2} \tau \]

6. **Final Result**:
Therefore, the speed of the particle at time \( t = \tau \) is:
\[ v(\tau) = \frac{b^2}{2} \tau \]