A particle moves with an initial velocity `V_(0)` and retardation `alpha v` , where `alpha` is a constant and v is the velocity at any time t.
total distance covered by the particle is

Options

A. `u//k`
B. uk
C. uk
D. `(u^(2))/( k )`

Answer

To solve the problem of finding the total distance covered by a particle moving with an initial velocity \( V_0 \) and a retardation proportional to its velocity, we can follow these steps:

### Step 1: Understand the Retardation
The problem states that the particle has a retardation given by \( \alpha v \), where \( \alpha \) is a constant and \( v \) is the velocity at any time \( t \). This means that the acceleration \( a \) can be expressed as:
\[ a = -\alpha v \]

### Step 2: Relate Acceleration to Velocity
Using the definition of acceleration, we can write:
\[ a = \frac{dv}{dt} \]
Thus, we have:
\[ \frac{dv}{dt} = -\alpha v \]

### Step 3: Separate Variables and Integrate
We can separate the variables to integrate:
\[ \frac{dv}{v} = -\alpha dt \]
Now, we integrate both sides. The limits for \( v \) will be from \( V_0 \) to \( v \) and for \( t \) from \( 0 \) to \( t \):
\[ \int_{V_0}^{v} \frac{1}{v} dv = -\alpha \int_{0}^{t} dt \]
This gives:
\[ \ln v - \ln V_0 = -\alpha t \]
Simplifying, we find:
\[ \ln \left(\frac{v}{V_0}\right) = -\alpha t \]
Exponentiating both sides, we obtain:
\[ v = V_0 e^{-\alpha t} \]

### Step 4: Relate Velocity to Distance
The velocity \( v \) is also defined as the rate of change of displacement \( s \):
\[ v = \frac{ds}{dt} \]
Substituting the expression for \( v \):
\[ \frac{ds}{dt} = V_0 e^{-\alpha t} \]

### Step 5: Integrate to Find Distance
Now, we can integrate to find the total distance \( s \):
\[ ds = V_0 e^{-\alpha t} dt \]
Integrating both sides, we have:
\[ s = \int_{0}^{t} V_0 e^{-\alpha t} dt \]
The integral of \( e^{-\alpha t} \) is:
\[ s = V_0 \left[-\frac{1}{\alpha} e^{-\alpha t}\right]_{0}^{t} \]
Evaluating this gives:
\[ s = -\frac{V_0}{\alpha} \left(e^{-\alpha t} - 1\right) = \frac{V_0}{\alpha} \left(1 - e^{-\alpha t}\right) \]

### Step 6: Find Total Distance as \( t \to \infty \)
To find the total distance covered by the particle, we take the limit as \( t \) approaches infinity:
\[ s = \frac{V_0}{\alpha} \left(1 - e^{-\alpha t}\right) \quad \text{as } t \to \infty \]
As \( t \to \infty \), \( e^{-\alpha t} \to 0 \):
\[ s = \frac{V_0}{\alpha} \left(1 - 0\right) = \frac{V_0}{\alpha} \]

### Final Answer
The total distance covered by the particle is:
\[ \boxed{\frac{V_0}{\alpha}} \]