Acceleration of a particle which is at rest at ` x=0 ` is `vec a = (4 -2 x) hat i`. 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. Maximum speed of particle is 2 units

Answer

To solve the problem step by step, we will analyze the given acceleration and find the required quantities.

### Step 1: Understanding the given acceleration
The acceleration of the particle is given by:
\[ \vec{a} = (4 - 2x) \hat{i} \]
This means that the acceleration depends on the position \(x\) of the particle.

### Step 2: Relating acceleration to velocity
We know that acceleration can be expressed in terms of velocity as:
\[ a = \frac{dv}{dt} = v \frac{dv}{dx} \]
Thus, we can set up the equation:
\[ v \frac{dv}{dx} = 4 - 2x \]

### Step 3: Separating variables and integrating
We can separate the variables and integrate:
\[ v \, dv = (4 - 2x) \, dx \]
Now, we integrate 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.

### Step 4: Finding the constant of integration
At \(x = 0\), the particle is at rest, so \(v = 0\):
\[ \frac{0^2}{2} = 4(0) - (0)^2 + C \implies C = 0 \]
Thus, the equation simplifies to:
\[ \frac{v^2}{2} = 4x - x^2 \]
or
\[ v^2 = 8x - 2x^2 \]

### Step 5: Finding points where the particle comes to rest
The particle comes to rest when \(v = 0\):
\[ 0 = 8x - 2x^2 \]
Factoring gives:
\[ 2x(4 - x) = 0 \]
This results in:
\[ x = 0 \quad \text{or} \quad x = 4 \]
Thus, the particle comes to rest at \(x = 0\) and \(x = 4\).

### Step 6: Analyzing oscillation
The acceleration can be rewritten as:
\[ \vec{a} = -2(x - 2) \hat{i} \]
This indicates that the particle experiences a restoring force towards \(x = 2\), suggesting simple harmonic motion (SHM) about \(x = 2\).

### Step 7: Finding maximum speed
To find the maximum speed, we need to find the maximum value of \(v\):
\[ v^2 = 8x - 2x^2 \]
To find the maximum, we differentiate with respect to \(x\):
\[ \frac{dv^2}{dx} = 8 - 4x \]
Setting this equal to zero gives:
\[ 8 - 4x = 0 \implies x = 2 \]
Now substituting \(x = 2\) back into the equation for \(v^2\):
\[ v^2 = 8(2) - 2(2^2) = 16 \]
Thus, \(v = \sqrt{16} = 4\).

### Final Results
1. The particle comes to rest at \(x = 0\) and \(x = 4\).
2. The particle oscillates about \(x = 2\).
3. The maximum speed of the particle is \(v = 4\).