A particle located at `x = 0` at time `t = 0`, starts moving along with the positive ` x-direction` with a velocity 'v' that varies as ` v = a sqrt(x)`. The displacement of the particle varies with time as

Options

A. ` t^(2)`
B. `t`
C. `t^(1/2)`
D. `t^(3)`

Explanation

To solve the problem, we need to find the displacement of a particle whose velocity varies with position as \( v = a \sqrt{x} \).

### Step-by-Step Solution:

1. **Understanding the relationship between velocity and displacement**:
The velocity \( v \) of the particle is given by:
\[ v = a \sqrt{x} \]
We know that velocity is also defined as the rate of change of displacement with respect to time:
\[ v = \frac{dx}{dt} \]
Therefore, we can equate the two expressions:
\[ \frac{dx}{dt} = a \sqrt{x} \]

2. **Separating variables**:
Rearranging the equation gives us:
\[ \frac{dx}{\sqrt{x}} = a \, dt \]

3. **Integrating both sides**:
We will integrate both sides. The left side requires the integral of \( \frac{1}{\sqrt{x}} \) and the right side is straightforward:
\[ \int \frac{dx}{\sqrt{x}} = \int a \, dt \]
The integral of \( \frac{1}{\sqrt{x}} \) is \( 2\sqrt{x} \), so we have:
\[ 2\sqrt{x} = at + C \]
where \( C \) is the constant of integration.

4. **Applying initial conditions**:
At \( t = 0 \), the particle is at \( x = 0 \). Plugging these values into the equation gives:
\[ 2\sqrt{0} = a(0) + C \implies C = 0 \]
Thus, our equation simplifies to:
\[ 2\sqrt{x} = at \]

5. **Solving for \( x \)**:
Now, we can solve for \( x \):
\[ \sqrt{x} = \frac{at}{2} \]
Squaring both sides gives:
\[ x = \left(\frac{at}{2}\right)^2 = \frac{a^2 t^2}{4} \]

6. **Final Result**:
Therefore, the displacement of the particle as a function of time is:
\[ x(t) = \frac{a^2}{4} t^2 \]

### Summary:
The displacement of the particle varies with time as \( x(t) = \frac{a^2}{4} t^2 \).

---