If the displacement of a particle varies with time as `sqrt x = t+ 3`

Options

A. velocity of the particle is inversely proportional to t
B. velocity of particle varies linearly with t
C. velocity of particle is proportional to `sqrt t`
D. initial velocity of the particle is zero

Explanation

To solve the problem, we need to analyze the given equation for displacement and find the velocity of the particle. Here are the steps to arrive at the solution:

### Step 1: Understand the given equation
The displacement of the particle is given by the equation:
\[ \sqrt{x} = t + 3 \]
We need to express \(x\) in terms of \(t\).

### Step 2: Square both sides
To eliminate the square root, we square both sides of the equation:
\[ x = (t + 3)^2 \]

### Step 3: Expand the equation
Now, we expand the right-hand side:
\[ x = t^2 + 6t + 9 \]

### Step 4: Differentiate to find velocity
The velocity \(v\) of the particle is defined as the rate of change of displacement with respect to time, which is given by:
\[ v = \frac{dx}{dt} \]
We differentiate \(x\) with respect to \(t\):
\[ v = \frac{d}{dt}(t^2 + 6t + 9) \]
Calculating the derivative:
\[ v = 2t + 6 \]

### Step 5: Analyze the velocity equation
The velocity equation \(v = 2t + 6\) is a linear function of \(t\). This means that the velocity varies linearly with time.

### Step 6: Identify the initial velocity
To find the initial velocity, we can substitute \(t = 0\) into the velocity equation:
\[ v(0) = 2(0) + 6 = 6 \]
Thus, the initial velocity of the particle is 6.

### Conclusion
From the analysis, we conclude that:
- The velocity of the particle varies linearly with time.
- The initial velocity of the particle is not zero; it is actually 6.

### Final Answer
The correct option is that the velocity of the particle varies linearly with \(t\) (Option 2).

---