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.
Velocity of particle at time is :

Options

A. `v = alpha t^(2)`
B. `v = V_(0) - alpha t`
C. `v = (V_(0))/(ln alpha t)`
D. `v = V_(0)e^(-alpha t)`

Explanation

To solve the problem, we need to find the velocity of a particle that starts with an initial velocity \( V_0 \) and experiences a retardation proportional to its velocity, given by \( \alpha v \), where \( \alpha \) is a constant.

### Step-by-Step Solution:

1. **Understanding Retardation**:
The retardation is given as \( \alpha v \). Since retardation is negative acceleration, we can express this as:
\[ a = -\alpha v \]

2. **Relating Acceleration to Velocity**:
We know that acceleration \( a \) can also be expressed as the time derivative of velocity:
\[ a = \frac{dv}{dt} \]
Therefore, we can write:
\[ \frac{dv}{dt} = -\alpha v \]

3. **Separating Variables**:
To solve this differential equation, we will separate the variables \( v \) and \( t \):
\[ \frac{dv}{v} = -\alpha dt \]

4. **Integrating Both Sides**:
Now we 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 us:
\[ \ln |v| = -\alpha t + C \]
where \( C \) is the constant of integration.

5. **Applying Initial Conditions**:
At \( t = 0 \), the velocity \( v = V_0 \). We can use this to find \( C \):
\[ \ln |V_0| = C \]
Therefore, we can rewrite our equation as:
\[ \ln |v| = -\alpha t + \ln |V_0| \]

6. **Exponentiating Both Sides**:
To solve for \( v \), we exponentiate both sides:
\[ |v| = e^{-\alpha t + \ln |V_0|} = |V_0| e^{-\alpha t} \]
Since velocity is positive, we can drop the absolute value:
\[ v = V_0 e^{-\alpha t} \]

### Final Answer:
The velocity of the particle at time \( t \) is:
\[ v(t) = V_0 e^{-\alpha t} \]