A particle starts moving in a straight line from rest with an acceleration ` ("in"m//s^(2) )` given by : ` a=12 sqrt x , ` where x is the distance travelled by the particle (in meters ) The velocity of the particle at the instant when its acceleration is ` 48m//s ^(2) ` is `.......m//s`

Explanation

To solve the problem, we need to find the velocity of a particle at the moment when its acceleration is 48 m/s², given that its acceleration is defined as \( a = 12 \sqrt{x} \), where \( x \) is the distance traveled by the particle in meters.

### Step-by-Step Solution:

1. **Set Up the Equation for Acceleration:**
We are given the acceleration as:
\[ a = 12 \sqrt{x} \]

2. **Substitute the Given Acceleration:**
We need to find the distance \( x \) when the acceleration \( a \) is 48 m/s². Set the equation equal to 48:
\[ 12 \sqrt{x} = 48 \]

3. **Solve for \( \sqrt{x} \):**
Divide both sides by 12:
\[ \sqrt{x} = \frac{48}{12} = 4 \]

4. **Square Both Sides to Find \( x \):**
\[ x = 4^2 = 16 \text{ m} \]

5. **Relate Acceleration to Velocity:**
We know that acceleration can also be expressed in terms of velocity and distance:
\[ a = v \frac{dv}{dx} \]
Substitute \( a = 12 \sqrt{x} \):
\[ 12 \sqrt{x} = v \frac{dv}{dx} \]

6. **Separate Variables:**
Rearranging gives:
\[ v \, dv = 12 \sqrt{x} \, dx \]

7. **Integrate Both Sides:**
Integrate the left side with respect to \( v \) and the right side with respect to \( x \):
\[ \int v \, dv = \int 12 \sqrt{x} \, dx \]
This results in:
\[ \frac{v^2}{2} = 12 \cdot \frac{2}{3} x^{3/2} + C \]
Simplifying gives:
\[ \frac{v^2}{2} = 8 x^{3/2} + C \]

8. **Determine the Constant of Integration \( C \):**
Since the particle starts from rest when \( x = 0 \) (initial velocity \( v = 0 \)):
\[ 0 = 8 \cdot 0^{3/2} + C \implies C = 0 \]
Thus, we have:
\[ \frac{v^2}{2} = 8 x^{3/2} \]

9. **Solve for Velocity \( v \):**
Multiply both sides by 2:
\[ v^2 = 16 x^{3/2} \]
Taking the square root gives:
\[ v = 4 x^{3/2} \]

10. **Substitute \( x = 16 \):**
Now substitute \( x = 16 \) to find \( v \):
\[ v = 4 \cdot (16)^{3/2} \]
Calculate \( (16)^{3/2} = (4^2)^{3/2} = 4^3 = 64 \):
\[ v = 4 \cdot 64 = 256 \text{ m/s} \]

### Final Answer:
The velocity of the particle at the instant when its acceleration is 48 m/s² is:
\[ \boxed{32} \text{ m/s} \]