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.
After how much time, speed of particle decreases by `75%`

Options

A. `(2l n2)/(alpha)`
B. `(l n(4//3))/(alpha)`
C. `(3V_(0))/(4alpha)`
D. `(e^(4))/(alpha)`

Explanation

To solve the problem step by step, we need to analyze the motion of the particle under the given conditions. Here's the detailed solution:

### Step 1: Understand the Problem
The particle starts with an initial velocity \( V_0 \) and experiences a retardation proportional to its velocity, expressed as \( -\alpha v \), where \( \alpha \) is a constant. We need to find the time \( t \) when the speed of the particle decreases to \( 25\% \) of its initial speed.

### Step 2: Set Up the Equation
The retardation can be expressed as:
\[ a = -\alpha v \]
Since acceleration \( a \) is the derivative of velocity \( v \) with respect to time \( t \), we can write:
\[ \frac{dv}{dt} = -\alpha v \]

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

### Step 4: Integrate Both Sides
Now we will integrate both sides. The left side will be integrated from the initial velocity \( V_0 \) to the final velocity \( 0.25 V_0 \), and the right side will be integrated from \( 0 \) to \( t \):
\[ \int_{V_0}^{0.25 V_0} \frac{1}{v} dv = \int_{0}^{t} -\alpha dt \]

### Step 5: Perform the Integration
The left side integrates to:
\[ \ln(v) \bigg|_{V_0}^{0.25 V_0} = \ln(0.25 V_0) - \ln(V_0) = \ln\left(\frac{0.25 V_0}{V_0}\right) = \ln(0.25) = \ln\left(\frac{1}{4}\right) \]
The right side integrates to:
\[ -\alpha t \]

### Step 6: Set the Integrated Results Equal
Setting the results from both sides equal gives:
\[ \ln\left(\frac{1}{4}\right) = -\alpha t \]

### Step 7: Solve for Time \( t \)
Rearranging for \( t \) gives:
\[ t = -\frac{\ln\left(\frac{1}{4}\right)}{\alpha} \]
Using the property of logarithms, we can simplify \( \ln\left(\frac{1}{4}\right) \):
\[ \ln\left(\frac{1}{4}\right) = \ln(4^{-1}) = -\ln(4) \]
Thus, we have:
\[ t = \frac{\ln(4)}{\alpha} \]

### Step 8: Further Simplification
Since \( \ln(4) = \ln(2^2) = 2\ln(2) \), we can express \( t \) as:
\[ t = \frac{2\ln(2)}{\alpha} \]

### Final Answer
The time after which the speed of the particle decreases by \( 75\% \) is:
\[ t = \frac{2\ln(2)}{\alpha} \]