A particle moves with initial velocity `v_(0)` and retardation `alphav`, where v is velocity at any instant t. Then the particle

Options

A. will cover a total distance `(V_(0))/(alpha)`
B. continues to move for a long time span.
C. attains a velocity `(v_(0)//2)` at `t = 1//alpha`
D. the particle comes to rest at `t = 1//alpha`

Answer

To solve the problem step by step, we will analyze the motion of the particle given its initial conditions and the nature of its retardation.

### Step 1: Understand the given parameters
The particle has:
- Initial velocity \( v_0 \)
- Retardation \( a = -\alpha v \), where \( v \) is the velocity at any instant \( t \).

### Step 2: Set up the equation for acceleration
From the definition of acceleration, we have:
\[ a = \frac{dv}{dt} \]
Substituting the expression for retardation:
\[ -\alpha v = \frac{dv}{dt} \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ -\alpha dt = \frac{dv}{v} \]

### Step 4: Integrate both sides
Integrate both sides. The left side will be integrated with respect to time \( t \) and the right side with respect to velocity \( v \):
\[ \int -\alpha dt = \int \frac{dv}{v} \]
This results in:
\[ -\alpha t = \ln v + C \]

### Step 5: Apply initial conditions to find constant \( C \)
At \( t = 0 \), \( v = v_0 \):
\[ -\alpha(0) = \ln v_0 + C \implies C = -\ln v_0 \]
Substituting \( C \) back into the equation gives:
\[ -\alpha t = \ln v - \ln v_0 \implies \ln \left(\frac{v}{v_0}\right) = -\alpha t \]

### Step 6: Solve for \( v \)
Exponentiating both sides results in:
\[ \frac{v}{v_0} = e^{-\alpha t} \implies v = v_0 e^{-\alpha t} \]

### Step 7: Determine when the particle comes to rest
The particle comes to rest when \( v = 0 \). As \( t \) approaches infinity, \( e^{-\alpha t} \) approaches 0, thus:
\[ \text{The particle stops moving as } t \to \infty. \]

### Step 8: Find the distance covered before coming to rest
Using the relationship \( a = \frac{dv}{dt} \) and substituting \( a = -\alpha v \):
\[ -\alpha v = v \frac{dv}{dx} \implies -\alpha dx = dv \]
Integrating from \( v_0 \) to \( 0 \) gives:
\[ -\alpha x = v \Big|_{v_0}^{0} \implies -\alpha x = 0 - v_0 \implies x = \frac{v_0}{\alpha} \]

### Conclusion
The total distance covered by the particle before it comes to rest is:
\[ x = \frac{v_0}{\alpha} \]

### Summary of Results
1. The particle comes to rest after covering a distance \( \frac{v_0}{\alpha} \).
2. The particle continues to move for a long time span until it eventually stops.