The relation between time and distance is is : ` a= ( 4-2x) ` .Select the correct alternative (s) .

Options

A. Particle further comes to rest at x = 4
B. Particle oscillates about x = 2
C. Maximum speed of particle is 4 units
D. All of the above

Explanation

To solve the problem, we start with the given relation between time and distance, which is \( a = 4 - 2x \). This equation describes the acceleration \( a \) of a particle as a function of its position \( x \).

### Step-by-Step Solution:

1. **Understanding the Relation**:
The equation \( a = 4 - 2x \) indicates that the acceleration of the particle depends on its position \( x \).

2. **Finding Velocity**:
We know that acceleration \( a \) can be expressed as the derivative of velocity \( v \) with respect to position \( x \):
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \]
Therefore, we can rewrite the equation as:
\[ v \frac{dv}{dx} = 4 - 2x \]

3. **Integrating to Find Velocity**:
We can separate variables and integrate:
\[ v dv = (4 - 2x) dx \]
Integrating both sides:
\[ \int v \, dv = \int (4 - 2x) \, dx \]
This gives us:
\[ \frac{v^2}{2} = 4x - x^2 + C \]
where \( C \) is the constant of integration.

4. **Finding the Maximum Speed**:
To find the maximum speed, we need to find the critical points of \( v \). We can do this by differentiating \( v^2 \) with respect to \( x \) and setting it to zero:
\[ \frac{d}{dx}(4x - x^2) = 4 - 2x = 0 \]
Solving for \( x \):
\[ 2x = 4 \implies x = 2 \]

5. **Calculating Maximum Speed**:
Substitute \( x = 2 \) back into the equation for \( v^2 \):
\[ v^2 = 2(4 \cdot 2 - 2^2) = 2(8 - 4) = 2 \cdot 4 = 8 \]
Thus, the maximum speed \( v \) is:
\[ v = \sqrt{8} = 2\sqrt{2} \]

6. **Determining Oscillation**:
The particle oscillates about \( x = 2 \) because the acceleration changes sign at this point, indicating that it is a point of equilibrium. The particle will oscillate between \( x = 0 \) and \( x = 4 \).

### Conclusion:
The correct statements are:
1. The maximum speed of the particle is \( 2\sqrt{2} \) at \( x = 2 \).
2. The particle oscillates about \( x = 2 \).