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 step by step, we will follow the reasoning provided in the video transcript, using the relationship between acceleration, velocity, and distance traveled by the particle.

### Step 1: Understand the given information
We have the acceleration of the particle given by:
\[ a = 12 \sqrt{x} \]
where \( x \) is the distance traveled in meters. The particle starts from rest, so its initial velocity \( u = 0 \).

### Step 2: Relate acceleration to velocity and distance
We can use the formula that relates acceleration to velocity and distance:
\[ a = v \frac{dv}{dx} \]
Substituting the expression for acceleration, we have:
\[ 12 \sqrt{x} = v \frac{dv}{dx} \]

### Step 3: Rearranging the equation
We can rearrange this equation to separate variables:
\[ v \, dv = 12 \sqrt{x} \, dx \]

### Step 4: Integrate both sides
Now we will integrate both sides. The left side integrates to:
\[ \int v \, dv = \frac{v^2}{2} + C_1 \]
And the right side integrates to:
\[ \int 12 \sqrt{x} \, dx = 12 \cdot \frac{2}{3} x^{3/2} + C_2 = 8 x^{3/2} + C_2 \]

Thus, we have:
\[ \frac{v^2}{2} = 8 x^{3/2} + C \]
where \( C = C_2 - C_1 \).

### Step 5: Determine the constant of integration
Since the particle starts from rest at \( x = 0 \) (where \( v = 0 \)), we can find \( C \):
\[ \frac{0^2}{2} = 8(0)^{3/2} + C \]
This implies \( C = 0 \).

### Step 6: Substitute back into the equation
Now we have:
\[ \frac{v^2}{2} = 8 x^{3/2} \]
Multiplying through by 2 gives:
\[ v^2 = 16 x^{3/2} \]

### Step 7: Find \( x \) when \( a = 48 \, \text{m/s}^2 \)
We need to find the distance \( x \) when the acceleration is \( 48 \, \text{m/s}^2 \):
\[ 48 = 12 \sqrt{x} \]
Dividing both sides by 12:
\[ \sqrt{x} = 4 \]
Squaring both sides gives:
\[ x = 16 \, \text{m} \]

### Step 8: Substitute \( x \) back to find \( v \)
Now we can substitute \( x = 16 \) back into the equation for \( v^2 \):
\[ v^2 = 16 (16)^{3/2} \]
Calculating \( (16)^{3/2} = 64 \):
\[ v^2 = 16 \cdot 64 = 1024 \]
Taking the square root gives:
\[ v = \sqrt{1024} = 32 \, \text{m/s} \]

### Final Answer
The velocity of the particle when its acceleration is \( 48 \, \text{m/s}^2 \) is:
\[ \boxed{32 \, \text{m/s}} \]