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 find the distance covered by a motorboat when its velocity decreases from \( v_0 \) to \( \frac{v_0}{2} \) after the engine is shut down. The resistance offered to the boat is proportional to the square of its velocity, given by \( \sigma v^2 \).

### Step-by-Step Solution:

1. **Understanding Forces**:
When the engine is shut down, the only force acting on the boat is the resistive force, which is given by \( F_{\text{resistance}} = \sigma v^2 \). According to Newton's second law, we have:
\[ F = ma = m \frac{dv}{dt} \]
Thus, we can write:
\[ m \frac{dv}{dt} = -\sigma v^2 \]
The negative sign indicates that the force is acting in the opposite direction of the motion.

2. **Changing Variables**:
We can express \( \frac{dv}{dt} \) in terms of displacement \( s \) using the chain rule:
\[ \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \frac{dv}{ds} \]
Therefore, substituting this into our equation gives:
\[ m v \frac{dv}{ds} = -\sigma v^2 \]

3. **Simplifying the Equation**:
We can cancel \( v \) from both sides (assuming \( v \neq 0 \)):
\[ m \frac{dv}{ds} = -\sigma v \]
Rearranging gives:
\[ \frac{dv}{ds} = -\frac{\sigma}{m} v \]

4. **Separating Variables**:
We can separate the variables to integrate:
\[ \frac{dv}{v} = -\frac{\sigma}{m} ds \]

5. **Integrating**:
Now we integrate both sides. The limits for \( v \) are from \( v_0 \) to \( \frac{v_0}{2} \) and for \( s \) from \( 0 \) to \( s \):
\[ \int_{v_0}^{\frac{v_0}{2}} \frac{1}{v} dv = -\frac{\sigma}{m} \int_{0}^{s} ds \]
This gives:
\[ \ln\left(\frac{v}{v_0}\right) \bigg|_{v_0}^{\frac{v_0}{2}} = -\frac{\sigma}{m} s \]
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:
\[ -\ln(2) = -\frac{\sigma}{m} s \]

6. **Solving 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:
\[ s = \frac{m \ln(2)}{\sigma} \]