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 reduce its speed to half of its initial value?

Options

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

Explanation

To solve the problem step by step, we will analyze the motion of the particle under the given conditions.

### Step 1: Understand the Retardation
The problem states that the particle experiences a retardation given by \( a = -\alpha v \), where \( v \) is the instantaneous velocity and \( \alpha \) is a constant. This means that the acceleration is acting in the opposite direction to the velocity.

### Step 2: Set Up the Differential Equation
We know that acceleration \( a \) can be expressed as the rate of change of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Substituting the expression for retardation, we have:
\[ \frac{dv}{dt} = -\alpha v \]

### Step 3: Rearrange the Equation
We can rearrange the equation to separate the variables:
\[ \frac{dv}{v} = -\alpha dt \]

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

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

### Step 6: Exponentiate to Solve for \( v \)
Exponentiating both sides gives:
\[ v = v_0 e^{-\alpha t} \]

### Step 7: Find the Time to Halve the Speed
We want to find the time \( t \) when the velocity \( v \) is half of the initial velocity \( v_0 \):
\[ \frac{v_0}{2} = v_0 e^{-\alpha t} \]
Dividing both sides by \( v_0 \) (assuming \( v_0 \neq 0 \)):
\[ \frac{1}{2} = e^{-\alpha t} \]

### Step 8: Take the Natural Logarithm
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{2}\right) = -\alpha t \]

### Step 9: Solve for \( t \)
Rearranging gives:
\[ t = -\frac{\ln\left(\frac{1}{2}\right)}{\alpha} \]
Using the property of logarithms, \( \ln\left(\frac{1}{2}\right) = -\ln(2) \):
\[ t = \frac{\ln(2)}{\alpha} \]

### Final Answer
The time taken for the particle to reduce its speed to half of its initial value is:
\[ t = \frac{\ln(2)}{\alpha} \]

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