A particle moves with an initial `v_(0)` and retardation alphav, where v is its velocity at any time t.
(i) The particle will cover a total distance `v_(0)/alpha`.
(ii) The particle will come to rest after time `1/alpha`.
(iii) The particle will continue to move for a very long time.
(iv) The velocity of the particle will become `v_(0)/2` after time `(1n2)/alpha`

Explanation

To solve the problem step by step, we will analyze the motion of the particle under the given conditions.

### Step 1: Understanding the Motion
The particle has an initial velocity \( v_0 \) and experiences a retardation proportional to its velocity, represented as \( -\alpha v \). This means that the acceleration \( a \) can be expressed as:
\[ a = -\alpha v \]

### Step 2: Setting Up the Differential Equation
We can express the relationship between velocity and time using the following differential equation:
\[ \frac{dv}{dt} = -\alpha v \]

### Step 3: Separating Variables
To solve this differential equation, we separate the variables:
\[ \frac{dv}{v} = -\alpha dt \]

### Step 4: Integrating Both Sides
Next, we integrate both sides:
\[ \int \frac{dv}{v} = \int -\alpha dt \]
This gives us:
\[ \ln |v| = -\alpha t + C \]
where \( C \) is the constant of integration.

### Step 5: Solving for the Constant
To find the constant \( C \), we use the initial condition \( v(0) = v_0 \):
\[ \ln |v_0| = C \]
Thus, we can rewrite the equation as:
\[ \ln |v| = -\alpha t + \ln |v_0| \]

### Step 6: Exponentiating to Find Velocity
Exponentiating both sides, we find:
\[ v = v_0 e^{-\alpha t} \]
This equation describes the velocity of the particle as a function of time.

### Step 7: Finding the Time to Come to Rest
To find when the particle comes to rest, we set \( v = 0 \). However, since \( e^{-\alpha t} \) never actually becomes zero, the particle will never come to rest in finite time. Thus, statement (ii) is incorrect.

### Step 8: Finding the Distance Traveled
To find the total distance traveled by the particle, we can use the relationship:
\[ s = \int_0^t v \, dt = \int_0^t v_0 e^{-\alpha t} \, dt \]
Calculating this integral:
\[ s = v_0 \int_0^t e^{-\alpha t} \, dt = v_0 \left[-\frac{1}{\alpha} e^{-\alpha t}\right]_0^t \]
Evaluating this gives:
\[ s = v_0 \left(-\frac{1}{\alpha} e^{-\alpha t} + \frac{1}{\alpha}\right) = \frac{v_0}{\alpha} (1 - e^{-\alpha t}) \]
As \( t \to \infty \), \( e^{-\alpha t} \to 0 \), thus:
\[ s \to \frac{v_0}{\alpha} \]
This confirms that statement (i) is correct.

### Step 9: Analyzing Long-Term Motion
As \( t \) approaches infinity, the velocity approaches zero, but the particle will continue to move for a very long time, confirming that statement (iii) is correct.

### Step 10: Finding the Time for Velocity to Become \( \frac{v_0}{2} \)
To find the time when the velocity becomes \( \frac{v_0}{2} \):
\[ \frac{v_0}{2} = v_0 e^{-\alpha t} \]
Dividing both sides by \( v_0 \) (assuming \( v_0 \neq 0 \)):
\[ \frac{1}{2} = e^{-\alpha t} \]
Taking the natural logarithm:
\[ -\alpha t = \ln \frac{1}{2} \]
Thus:
\[ t = -\frac{\ln 2}{\alpha} \]
This confirms that statement (iv) is incorrect since it states \( \frac{\ln 2}{\alpha} \).

### Summary of Correct Statements
- (i) The particle will cover a total distance \( \frac{v_0}{\alpha} \) - **Correct**
- (ii) The particle will come to rest after time \( \frac{1}{\alpha} \) - **Incorrect**
- (iii) The particle will continue to move for a very long time - **Correct**
- (iv) The velocity of the particle will become \( \frac{v_0}{2} \) after time \( \frac{\ln 2}{\alpha} \) - **Incorrect**