Acceleration of particle moving rectilinearly is `a = 4-2x`(where x is position in metre and a in ms^(-2))`. It si at instantaneous rest at x = 0. At what position `x 9`in meter) will the particle again come to instantaneous rest?

Explanation

To solve the problem, we need to find the position \( x \) where the particle comes to instantaneous rest again, given the acceleration function \( a = 4 - 2x \).

### Step-by-Step Solution:

1. **Understand the relationship between acceleration, velocity, and position**:
The acceleration \( a \) can be expressed in terms of velocity \( v \) and position \( x \) using the chain rule:
\[ a = v \frac{dv}{dx} \]
Therefore, we can set up the equation:
\[ v \frac{dv}{dx} = 4 - 2x \]

2. **Separate variables**:
Rearranging gives us:
\[ v \, dv = (4 - 2x) \, dx \]

3. **Integrate both sides**:
We will integrate the left side with respect to \( v \) and the right side with respect to \( x \):
\[ \int v \, dv = \int (4 - 2x) \, dx \]
The left side integrates to:
\[ \frac{v^2}{2} \]
The right side integrates to:
\[ 4x - x^2 + C \]
where \( C \) is the constant of integration.

4. **Combine the results**:
Thus, we have:
\[ \frac{v^2}{2} = 4x - x^2 + C \]

5. **Determine the constant \( C \)**:
We know that at \( x = 0 \), the particle is at instantaneous rest, which means \( v = 0 \). Substituting these values into the equation:
\[ 0 = 4(0) - (0)^2 + C \implies C = 0 \]
Therefore, the equation simplifies to:
\[ \frac{v^2}{2} = 4x - x^2 \]

6. **Find when the particle comes to rest again**:
The particle comes to rest when \( v = 0 \):
\[ 0 = 4x - x^2 \]
Rearranging gives:
\[ x^2 - 4x = 0 \]
Factoring out \( x \):
\[ x(x - 4) = 0 \]
This gives us two solutions:
\[ x = 0 \quad \text{or} \quad x = 4 \]

7. **Conclusion**:
The particle comes to instantaneous rest again at \( x = 4 \) meters.

### Final Answer:
The particle will again come to instantaneous rest at \( x = 4 \) meters.