A particle starts moving in a straight line with initial velocity `v_(0)` Applied forces cases a retardation of av, where v is magnitude of instantaneous velocity and `alpha` is a constant.
Total distance covered by the particle is

Options

A. infinity
B. `v_(0)alpha`
C. `v_(0)//alpha`
D. `v_(0)//ealpha`

Explanation

To solve the problem of a particle moving in a straight line with an initial velocity \( v_0 \) and experiencing a retardation proportional to its instantaneous velocity, we can follow these steps:

### Step 1: Understand the Retardation
The retardation (deceleration) is given as \( a = -\alpha v \), where \( \alpha \) is a constant and \( v \) is the instantaneous velocity of the particle. This indicates that the acceleration is in the opposite direction to the velocity.

### Step 2: Set Up the Differential Equation
We can express the acceleration as the derivative of velocity with respect to time:
\[ \frac{dv}{dt} = -\alpha v \]
This is a separable differential equation.

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

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

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

### Step 6: Exponentiate to Solve for Velocity
Exponentiating both sides gives:
\[ v = v_0 e^{-\alpha t} \]
This shows that the velocity decreases exponentially over time.

### Step 7: Find the Distance Covered
The distance \( s \) covered by the particle can be expressed as:
\[ s = \int v \, dt \]
Substituting for \( v \):
\[ s = \int v_0 e^{-\alpha t} \, dt \]

### Step 8: Integrate to Find Distance
The integral becomes:
\[ s = v_0 \int e^{-\alpha t} \, dt = v_0 \left(-\frac{1}{\alpha} e^{-\alpha t}\right) + C \]
Applying the limits from \( t = 0 \) to \( t = t \):
\[ s = -\frac{v_0}{\alpha} e^{-\alpha t} \bigg|_0^t = -\frac{v_0}{\alpha} (e^{-\alpha t} - 1) \]
This simplifies to:
\[ s = \frac{v_0}{\alpha} (1 - e^{-\alpha t}) \]

### Step 9: Determine the Total Distance Before Stopping
As \( t \) approaches infinity, \( e^{-\alpha t} \) approaches 0:
\[ s = \frac{v_0}{\alpha} (1 - 0) = \frac{v_0}{\alpha} \]

### Final Answer
The total distance covered by the particle before it stops is:
\[ \boxed{\frac{v_0}{\alpha}} \]