A motor boat of mass m moves along a lake with velocity `v_(0)`. At t = 0, the engine of the boat is shut down. Resistance offered to the boat is equal to `sigma v^(2)`. Then distance covered by the boat when its velocity becomes `v_(0)//2` is

Options

A. `(3mv_(0))/(2sigma)`
B. `(v_(0)m ln2)/(2sigma)`
C. `(2m In2)/(sigma)`
D. `(m In2)/(sigma)`

Answer

To solve the problem, we need to analyze the motion of the motorboat after the engine is shut down. The resistance acting on the boat is proportional to the square of its velocity. Let's break down the solution step by step.

### Step 1: Understand the Forces Acting on the Boat
When the engine is shut down, the only force acting on the boat is the resistance force, which is given by:
\[ R = \sigma v^2 \]
where \( R \) is the resistance, \( \sigma \) is a constant, and \( v \) is the velocity of the boat.

### Step 2: Apply Newton's Second Law
According to Newton's second law, the net force acting on the boat is equal to the mass of the boat multiplied by its acceleration:
\[ m a = -R \]
Since the resistance is acting in the opposite direction to the motion, we can write:
\[ m a = -\sigma v^2 \]

### Step 3: Relate Acceleration to Velocity
Acceleration \( a \) can be expressed in terms of velocity \( v \) and position \( x \) using the chain rule:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \]
Substituting this into the equation gives:
\[ m v \frac{dv}{dx} = -\sigma v^2 \]

### Step 4: Simplify the Equation
We can simplify the equation by dividing both sides by \( v \) (assuming \( v \neq 0 \)):
\[ m \frac{dv}{dx} = -\sigma v \]

### Step 5: Rearranging the Equation
Rearranging gives us:
\[ \frac{dv}{v} = -\frac{\sigma}{m} dx \]

### Step 6: Integrate Both Sides
We need to integrate both sides. The left side will be integrated from the initial velocity \( v_0 \) to the final velocity \( \frac{v_0}{2} \), and the right side will be integrated from \( 0 \) to \( s \) (the distance covered):
\[ \int_{v_0}^{\frac{v_0}{2}} \frac{dv}{v} = -\frac{\sigma}{m} \int_{0}^{s} dx \]
This results in:
\[ \ln\left(\frac{v}{v_0}\right) \bigg|_{v_0}^{\frac{v_0}{2}} = -\frac{\sigma}{m} s \]

### Step 7: Evaluate the Integrals
Evaluating the left side:
\[ \ln\left(\frac{\frac{v_0}{2}}{v_0}\right) = \ln\left(\frac{1}{2}\right) = -\ln(2) \]
Thus, we have:
\[ -\ln(2) = -\frac{\sigma}{m} s \]

### Step 8: Solve for Distance \( s \)
Rearranging gives:
\[ s = \frac{m \ln(2)}{\sigma} \]

### Final Answer
The distance covered by the boat when its velocity becomes \( \frac{v_0}{2} \) is:
\[ s = \frac{m \ln(2)}{\sigma} \]