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.
How long the particle will take to come to rest.

Options

A. `("in"2)/(alpha)`
B. `alpha` In 2
C. `(1)/(alpha)`
D. infinite

Explanation

To solve the problem, we need to analyze the motion of a particle that starts with an initial velocity \( v_0 \) and experiences a retardation proportional to its instantaneous velocity. The retardation is given as \( -\alpha v \), where \( \alpha \) is a constant.

### Step-by-Step Solution:

1. **Understanding the Retardation**:
The retardation (negative acceleration) acting on the particle is given by:
\[ a = -\alpha v \]
Here, \( v \) is the instantaneous velocity of the particle.

2. **Using the Definition of Acceleration**:
We know that acceleration is the rate of change of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Therefore, we can write:
\[ \frac{dv}{dt} = -\alpha v \]

3. **Separating Variables**:
We can rearrange the equation to separate the variables \( v \) and \( t \):
\[ \frac{dv}{v} = -\alpha dt \]

4. **Integrating Both Sides**:
We will integrate both sides. The left side will be integrated with respect to \( v \) and the right side with respect to \( t \):
\[ \int \frac{dv}{v} = \int -\alpha dt \]
This gives:
\[ \ln |v| = -\alpha t + C \]
where \( C \) is the constant of integration.

5. **Applying Initial Conditions**:
At \( t = 0 \), the velocity \( v = v_0 \):
\[ \ln |v_0| = C \]
Thus, we can substitute \( C \) back into the equation:
\[ \ln |v| = -\alpha t + \ln |v_0| \]

6. **Exponentiating Both Sides**:
To eliminate the natural logarithm, we exponentiate both sides:
\[ |v| = |v_0| e^{-\alpha t} \]

7. **Finding the Time to Come to Rest**:
The particle comes to rest when \( v = 0 \). Setting \( |v| = 0 \):
\[ 0 = v_0 e^{-\alpha t} \]
Since \( v_0 \) is a positive constant, the only way for the equation to hold true is if \( e^{-\alpha t} \) approaches zero. This occurs as \( t \) approaches infinity:
\[ t \to \infty \]

### Conclusion:
The time taken for the particle to come to rest is infinite.