The relation `3t=sqrt(3x)+6` describe the displacement of a particle in one direction where x is in metres and t in sec.
The displacement, when velocity is zero is

Options

A. 24 metres
B. 12 metres
C. 5 metres
D. zero

Explanation

To solve the problem step by step, we need to find the displacement of a particle when its velocity is zero, given the relation:

\[ 3t = \sqrt{3x} + 6 \]

### Step 1: Rearranging the equation
First, we will isolate \( \sqrt{3x} \) in the equation.

\[ 3t - 6 = \sqrt{3x} \]

### Step 2: Squaring both sides
To eliminate the square root, we will square both sides of the equation.

\[ (3t - 6)^2 = 3x \]

### Step 3: Expanding the left side
Now, we will expand the left side of the equation.

\[ (3t - 6)(3t - 6) = 9t^2 - 36t + 36 \]

Thus, we have:

\[ 9t^2 - 36t + 36 = 3x \]

### Step 4: Solving for \( x \)
Now, we can express \( x \) in terms of \( t \).

\[ x = \frac{9t^2 - 36t + 36}{3} \]

Simplifying this gives:

\[ x = 3t^2 - 12t + 12 \]

### Step 5: Finding the velocity
The velocity \( v \) is defined as the derivative of displacement with respect to time, \( \frac{dx}{dt} \).

\[ v = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - 12t + 12) \]

Calculating the derivative:

\[ v = 6t - 12 \]

### Step 6: Setting the velocity to zero
To find when the velocity is zero, we set \( v = 0 \):

\[ 6t - 12 = 0 \]

Solving for \( t \):

\[ 6t = 12 \quad \Rightarrow \quad t = 2 \text{ seconds} \]

### Step 7: Finding displacement at \( t = 2 \)
Now we will substitute \( t = 2 \) back into the displacement equation to find \( x \):

\[ x = 3(2)^2 - 12(2) + 12 \]

Calculating this gives:

\[ x = 3(4) - 24 + 12 \]
\[ x = 12 - 24 + 12 \]
\[ x = 0 \]

### Final Answer
Thus, the displacement when the velocity is zero is:

\[ \boxed{0 \text{ meters}} \]