A particle starts from point `A`, moves along a straight line path with an accleration given by `a = 2 (4 - x)` where `x` is distance from point `A`. The particle stops at point `B` for a moment. Find the distance `AB` (in `m`). (All values are in `S.I.` units)

Explanation

To solve the problem, we need to find the distance `AB` where a particle starts from point `A` and stops at point `B` under the influence of a given acceleration. The acceleration is defined as:

\[ a = 2(4 - x) \]

where \( x \) is the distance from point `A`.

### Step-by-Step Solution:

1. **Understand the acceleration equation**:
The acceleration can be rewritten as:
\[ a = 8 - 2x \]

2. **Relate acceleration to velocity**:
We know that acceleration \( a \) can be expressed as:
\[ a = \frac{dv}{dt} = v \frac{dv}{dx} \]
Therefore, we can set up the equation:
\[ v \frac{dv}{dx} = 8 - 2x \]

3. **Separate variables and integrate**:
Rearranging gives us:
\[ v dv = (8 - 2x) dx \]
Now we integrate both sides. The left side integrates to:
\[ \int v \, dv = \frac{v^2}{2} \]
The right side integrates to:
\[ \int (8 - 2x) \, dx = 8x - x^2 \]
Thus, we have:
\[ \frac{v^2}{2} = 8x - x^2 + C \]

4. **Apply initial conditions**:
At point `A`, when \( x = 0 \), the particle starts from rest, so \( v = 0 \). Plugging in these values gives:
\[ 0 = 8(0) - (0)^2 + C \implies C = 0 \]
Therefore, our equation simplifies to:
\[ \frac{v^2}{2} = 8x - x^2 \]

5. **Determine when the particle stops**:
The particle stops at point `B`, meaning \( v = 0 \). Setting \( v = 0 \) in our equation gives:
\[ 0 = 8x - x^2 \]
Rearranging this, we factor out \( x \):
\[ x(8 - x) = 0 \]
This gives us two solutions: \( x = 0 \) (point A) or \( x = 8 \).

6. **Conclusion**:
The distance \( AB \) is therefore:
\[ AB = 8 \, \text{m} \]

### Final Answer:
The distance \( AB \) is \( 8 \, \text{m} \).