A particle is moving along `x`-axis. Its velocity `v` with `x` co-ordinate is varying as `v=sqrt(x)`. Then

Options

A. initial velocity of particle is zero
B. motion is non-uniformly accelerated
C. acceleration of particle at `x=2m` is `1/2m//s^(2)`
D. acceleration of particle at `x=4m` is `1m//s^(2)`

Answer

To solve the problem step by step, we will analyze the given information and derive the necessary conclusions.

### Step 1: Understand the relationship between velocity and position
The velocity of the particle is given by the equation:
\[ v = \sqrt{x} \]
This indicates that the velocity depends on the position \( x \).

### Step 2: Determine the initial velocity
When \( x = 0 \):
\[ v = \sqrt{0} = 0 \]
Thus, the initial velocity of the particle is zero.

### Step 3: Find the acceleration
Acceleration \( a \) can be expressed in terms of velocity and position using the chain rule:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \cdot \frac{dv}{dx} \]
Substituting \( v = \sqrt{x} \):
\[ a = \sqrt{x} \cdot \frac{dv}{dx} \]

### Step 4: Calculate \( \frac{dv}{dx} \)
To find \( \frac{dv}{dx} \), we differentiate \( v = \sqrt{x} \):
\[ \frac{dv}{dx} = \frac{1}{2\sqrt{x}} \]

### Step 5: Substitute \( \frac{dv}{dx} \) back into the acceleration equation
Now substituting \( \frac{dv}{dx} \) into the acceleration equation:
\[ a = \sqrt{x} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2} \text{ m/s}^2 \]

### Step 6: Analyze the motion
Since the acceleration \( a = \frac{1}{2} \text{ m/s}^2 \) is constant, the motion is uniformly accelerated.

### Step 7: Calculate acceleration at specific positions
- At \( x = 2 \) m:
\[ a = \frac{1}{2} \text{ m/s}^2 \]
- At \( x = 4 \) m:
\[ a = \frac{1}{2} \text{ m/s}^2 \]

### Conclusion
1. The initial velocity of the particle is zero.
2. The motion is uniformly accelerated.
3. The acceleration of the particle at \( x = 2 \) m is \( \frac{1}{2} \text{ m/s}^2 \).
4. The acceleration of the particle at \( x = 4 \) m is also \( \frac{1}{2} \text{ m/s}^2 \), not \( 1 \text{ m/s}^2 \).

### Summary of Options
- Option A: Correct (initial velocity is zero)
- Option B: Incorrect (motion is uniformly accelerated)
- Option C: Correct (acceleration at \( x = 2 \) m is \( \frac{1}{2} \text{ m/s}^2 \))
- Option D: Incorrect (acceleration at \( x = 4 \) m is \( \frac{1}{2} \text{ m/s}^2 \))