A particle moves with initial velocity `v_(0)` and retardation `alphav`, where v is velocity at any instant t. Then the particle

Options

A. will cover a total distance `(V_(0))/(alpha)`
B. continues to move for a long time span.
C. attains a velocity `(v_(0)//2)` at `t = 1//alpha`
D. the particle comes to rest at `t = 1//alpha`

Answer

To solve the problem of 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 relationship between acceleration and velocity
The retardation (deceleration) is given as \( -\alpha v \), where \( \alpha \) is a constant. This means that the acceleration \( a \) can be expressed as:
\[ a = \frac{dv}{dt} = -\alpha v \]

### Step 2: Rearranging the equation
We can rearrange the equation to separate variables:
\[ \frac{dv}{v} = -\alpha dt \]

### Step 3: Integrate both sides
Integrate both sides. The left side will be integrated from \( v_0 \) to \( v \) and the right side from \( 0 \) to \( t \):
\[ \int_{v_0}^{v} \frac{dv}{v} = -\alpha \int_{0}^{t} dt \]
This gives:
\[ \ln v - \ln v_0 = -\alpha t \]
or
\[ \ln \left(\frac{v}{v_0}\right) = -\alpha t \]

### Step 4: Solve for \( v \)
Exponentiating both sides, we find:
\[ \frac{v}{v_0} = e^{-\alpha t} \]
Thus, we can express \( v \) as:
\[ v = v_0 e^{-\alpha t} \]

### Step 5: Determine when the particle comes to rest
The particle will come to rest when \( v = 0 \). However, from the equation \( v = v_0 e^{-\alpha t} \), we see that \( v \) approaches zero as \( t \) approaches infinity. Therefore, the particle never actually comes to rest in a finite amount of time.

### Step 6: Find the time when the velocity is half of the initial velocity
Set \( v = \frac{v_0}{2} \):
\[ \frac{v_0}{2} = v_0 e^{-\alpha t} \]
Dividing both sides by \( v_0 \) (assuming \( v_0 \neq 0 \)):
\[ \frac{1}{2} = e^{-\alpha t} \]
Taking the natural logarithm:
\[ -\alpha t = \ln \left(\frac{1}{2}\right) \]
Thus,
\[ t = \frac{\ln 2}{\alpha} \]

### Step 7: Calculate the distance traveled
To find the distance \( x \) traveled, we use the relationship \( v = \frac{dx}{dt} \):
\[ \frac{dx}{dt} = v_0 e^{-\alpha t} \]
Integrating both sides gives:
\[ dx = v_0 e^{-\alpha t} dt \]
Integrating from \( 0 \) to \( t \):
\[ x = \int_0^t v_0 e^{-\alpha t} dt = \left[-\frac{v_0}{\alpha} e^{-\alpha t}\right]_0^t \]
Evaluating the integral:
\[ x = -\frac{v_0}{\alpha} e^{-\alpha t} + \frac{v_0}{\alpha} \]
This simplifies to:
\[ x = \frac{v_0}{\alpha} (1 - e^{-\alpha t}) \]

### Step 8: Find the total distance when the particle stops
As \( t \) approaches infinity, \( e^{-\alpha t} \) approaches 0, and the total distance \( x \) becomes:
\[ x = \frac{v_0}{\alpha} \]

### Conclusion
Thus, the particle travels a total distance of \( \frac{v_0}{\alpha} \) before it effectively comes to rest.