If the velocity of the particle is given by `v=sqrt(x)` and initially particle was at `x=4m` then which of the following are correct.

Options

A. At `t=2 s`, the position of the particle is at `x=9m`
B. Particle acceleration at `t=2 s` is `1 m//s^(2)`
C. Particle acceleration is `1//2 m//s^(2)` through out the motion
D. Particle will never go in negative direction from it's starting position

Explanation

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

### Step 1: Understand the relationship between velocity and position
The velocity of the particle is given by the equation:
\[ v = \sqrt{x} \]
We know that velocity \( v \) can also be expressed as:
\[ v = \frac{dx}{dt} \]
Thus, we can equate the two expressions:
\[ \frac{dx}{dt} = \sqrt{x} \]

### Step 2: Rearrange the equation for integration
Rearranging the equation gives:
\[ dx = \sqrt{x} \, dt \]
To facilitate integration, we can express this as:
\[ \frac{dx}{\sqrt{x}} = dt \]

### Step 3: Integrate both sides
Now, we will integrate both sides:
- The left side will be integrated from the initial position \( x = 4 \) to \( x \).
- The right side will be integrated from \( t = 0 \) to \( t \).

The integration gives:
\[ \int_{4}^{x} \frac{dx}{\sqrt{x}} = \int_{0}^{t} dt \]

Calculating the integrals:
- The left side:
\[ 2\sqrt{x} \bigg|_{4}^{x} = 2\sqrt{x} - 4 \]
- The right side:
\[ t - 0 = t \]

Setting these equal gives us:
\[ 2\sqrt{x} - 4 = t \]

### Step 4: Solve for \( x \)
Rearranging the equation:
\[ 2\sqrt{x} = t + 4 \]
\[ \sqrt{x} = \frac{t + 4}{2} \]
Squaring both sides yields:
\[ x = \left(\frac{t + 4}{2}\right)^2 \]

### Step 5: Substitute \( t = 2 \) seconds to find \( x \)
Now, we can substitute \( t = 2 \) seconds into the equation:
\[ x = \left(\frac{2 + 4}{2}\right)^2 = \left(\frac{6}{2}\right)^2 = 3^2 = 9 \, \text{meters} \]

### Step 6: Determine the acceleration
To find the acceleration, we first need to find the velocity as a function of time:
From the expression for \( x \):
\[ x = \left(\frac{t + 4}{2}\right)^2 \]
Differentiating \( x \) with respect to \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt}\left(\frac{t + 4}{2}\right)^2 = \frac{1}{2} \cdot 2\left(\frac{t + 4}{2}\right) \cdot \frac{1}{2} = \frac{t + 4}{2} \]

Now, differentiating \( v \) with respect to \( t \) to find acceleration:
\[ a = \frac{dv}{dt} = \frac{1}{2} \]

### Step 7: Analyze the results
1. At \( t = 2 \) seconds, the position \( x = 9 \) meters (Correct).
2. The acceleration \( a = \frac{1}{2} \, \text{m/s}^2 \) (Incorrect statement in the question).
3. The acceleration is constant at \( \frac{1}{2} \, \text{m/s}^2 \) (Correct).
4. The particle will never go in the negative direction from its starting point (Correct).

### Final Answers:
- The correct options are:
- 1st option: Correct (9 meters at \( t = 2 \) seconds).
- 2nd option: Incorrect (acceleration is \( \frac{1}{2} \, \text{m/s}^2 \)).
- 3rd option: Correct (acceleration is \( \frac{1}{2} \, \text{m/s}^2 \)).
- 4th option: Correct (particle never goes negative).