The relation between acceleration 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 need to analyze the given relation between acceleration and distance, which is expressed as:

\[ a = 4 - 2x \]

### Step 1: Understanding the Relation
The first step is to understand what this equation implies. Here, \( a \) is the acceleration of the particle, and \( x \) is the distance. The equation indicates that the acceleration is dependent on the position \( x \).

### Step 2: Finding Velocity
To find the velocity of the particle, we can use the relation between acceleration and velocity. We know that:

\[ a = v \frac{dv}{dx} \]

Substituting the expression for acceleration into this equation gives:

\[ v \frac{dv}{dx} = 4 - 2x \]

### Step 3: Separating Variables
Next, we separate the variables to integrate:

\[ v \, dv = (4 - 2x) \, dx \]

### Step 4: Integrating Both Sides
Now, we integrate both sides:

\[ \int v \, dv = \int (4 - 2x) \, dx \]

The left side integrates to:

\[ \frac{v^2}{2} \]

The right side integrates to:

\[ 4x - x^2 + C \]

Thus, we have:

\[ \frac{v^2}{2} = 4x - x^2 + C \]

### Step 5: Solving for Velocity
Multiplying through by 2 gives:

\[ v^2 = 8x - 2x^2 + 2C \]

Taking the square root:

\[ v = \pm \sqrt{8x - 2x^2 + 2C} \]

### Step 6: Finding Conditions for Rest
To determine if the particle comes to rest at \( x = 4 \), we set \( v = 0 \):

\[ 0 = 8(4) - 2(4^2) + 2C \]

This simplifies to:

\[ 0 = 32 - 32 + 2C \implies 2C = 0 \implies C = 0 \]

Thus, the velocity equation simplifies to:

\[ v = \sqrt{8x - 2x^2} \]

### Step 7: Verifying Rest Condition
Now, substituting \( x = 4 \):

\[ v = \sqrt{8(4) - 2(4^2)} = \sqrt{32 - 32} = 0 \]

This confirms that the particle comes to rest at \( x = 4 \).

### Step 8: Finding Maximum Speed
To find the maximum speed, we differentiate the velocity equation with respect to \( x \) and set it to zero:

\[ \frac{dv}{dx} = \frac{1}{2\sqrt{8x - 2x^2}} \cdot (8 - 4x) = 0 \]

Setting the numerator to zero gives:

\[ 8 - 4x = 0 \implies x = 2 \]

### Step 9: Maximum Speed Calculation
Now, substituting \( x = 2 \) back into the velocity equation:

\[ v = \sqrt{8(2) - 2(2^2)} = \sqrt{16 - 8} = \sqrt{8} = 2\sqrt{2} \]

### Step 10: Conclusion
Thus, the maximum speed of the particle is \( 2\sqrt{2} \), which is approximately \( 2.83 \), and not 4 units.

### Final Verification
1. The particle comes to rest at \( x = 4 \) (True).
2. The maximum speed is \( 2\sqrt{2} \) (False).
3. The particle oscillates about \( x = 2 \) (True).

### Correct Alternatives
The correct alternatives are:
- The particle comes to rest at \( x = 4 \).
- The particle oscillates about \( x = 2 \).