A body is moving down along inclined plane of angle of inclination q. The coefficient of friction between the body and the plane varies as mu = 0.5x, where x is the distance moved down the plane. The body will have the maximum velocity when it has travelled a distance x given by 

Options

A. `x = 2 tan theta`
B. `x = 2/(tan theta)`
C. `x = sqrt(2) cot theta`
D. `x = (sqrt(2))/(cot theta)`

Answer

To solve the problem, we need to analyze the forces acting on the body moving down the inclined plane and find the distance \( x \) at which the body has maximum velocity.

### Step-by-Step Solution:

1. **Identify the Forces:**
The forces acting on the body include:
- Gravitational force (\( mg \)) acting vertically downward.
- Normal force (\( N \)) acting perpendicular to the inclined plane.
- Frictional force (\( f \)) acting opposite to the direction of motion.

2. **Resolve the Gravitational Force:**
The gravitational force can be resolved into two components:
- Parallel to the incline: \( mg \sin \theta \)
- Perpendicular to the incline: \( mg \cos \theta \)

3. **Determine the Normal Force:**
The normal force \( N \) is balanced by the perpendicular component of the gravitational force:
\[ N = mg \cos \theta \]

4. **Express the Frictional Force:**
The frictional force \( f \) can be expressed using the coefficient of friction \( \mu \):
\[ f = \mu N = \mu (mg \cos \theta) \]
Given that \( \mu = 0.5x \), we can substitute this into the equation:
\[ f = (0.5x)(mg \cos \theta) \]

5. **Set Up the Equation for Maximum Velocity:**
For maximum velocity, the net force acting down the incline should be zero, meaning the frictional force equals the component of gravitational force down the incline:
\[ mg \sin \theta = f \]
Substituting the expression for \( f \):
\[ mg \sin \theta = (0.5x)(mg \cos \theta) \]

6. **Cancel \( mg \) from Both Sides:**
Since \( mg \) is present on both sides of the equation, we can cancel it out:
\[ \sin \theta = 0.5x \cos \theta \]

7. **Rearranging the Equation:**
Rearranging gives:
\[ x = \frac{\sin \theta}{0.5 \cos \theta} \]

8. **Simplify the Expression:**
Recognizing that \( 0.5 \) is equivalent to \( \frac{1}{2} \):
\[ x = \frac{\sin \theta}{\frac{1}{2} \cos \theta} = 2 \frac{\sin \theta}{\cos \theta} = 2 \tan \theta \]

9. **Final Result:**
Therefore, the distance \( x \) at which the body will have the maximum velocity is:
\[ x = 2 \tan \theta \]