Velocityof a particle as function of displacement `x` is given by `v=bx^(1//2)`. Then the displacement as function of time is:

Options

A. `bt`
B. `(b^(2)t^(2))/(4)`
C. `(bt)/(4)`
D. `(b^(2)t^(3))/(4)`

Answer

To find the displacement as a function of time given the velocity as a function of displacement, we can follow these steps:

### Step-by-Step Solution:

1. **Start with the given velocity equation**:
\[ v = bx^{1/2} \]
where \( b \) is a constant.

2. **Relate velocity to displacement**:
Velocity \( v \) can also be expressed as the derivative of displacement \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} \]
Therefore, we can set the two expressions for velocity equal to each other:
\[ \frac{dx}{dt} = bx^{1/2} \]

3. **Rearrange the equation**:
To separate variables, we can rearrange the equation:
\[ \frac{dx}{x^{1/2}} = b \, dt \]

4. **Integrate both sides**:
Now we will integrate both sides. The left side requires the integration of \( x^{-1/2} \):
\[ \int x^{-1/2} \, dx = \int b \, dt \]
The integral of \( x^{-1/2} \) is:
\[ 2x^{1/2} = bt + C \]
where \( C \) is the constant of integration.

5. **Solve for \( x \)**:
To isolate \( x \), we can rearrange the equation:
\[ x^{1/2} = \frac{bt + C}{2} \]
Squaring both sides gives:
\[ x = \left(\frac{bt + C}{2}\right)^2 \]

6. **Simplify the expression**:
Expanding the squared term:
\[ x = \frac{(bt + C)^2}{4} \]

### Final Result:
Thus, the displacement \( x \) as a function of time \( t \) is given by:
\[ x = \frac{(bt + C)^2}{4} \]