A car is negotiating a curved road of radius R. The road is banked at angle `theta`. The coefficeint of friction between the tyres of the car and the road is `mu_(s)`. The maximum safe velocity on this road is

Options

A. `4.5`m/w
B. `5.0` m/s
C. `5.7` m/s
D. `6.2`m/s

Explanation

To find the maximum safe velocity of a car negotiating a banked curve, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Forces Acting on the Car**:
- When a car is negotiating a banked curve, the forces acting on it include gravitational force (mg), normal force (N), and frictional force (f).
- The gravitational force acts vertically downward, while the normal force acts perpendicular to the surface of the road.

2. **Setting Up the Equations**:
- The car moves in a circular path, so we can analyze the forces in both horizontal and vertical directions.
- In the horizontal direction, the net force provides the centripetal force required for circular motion:
\[ N \sin \theta + f \cos \theta = \frac{mv^2}{R} \]
- In the vertical direction, the forces must balance:
\[ N \cos \theta = mg + f \sin \theta \]

3. **Expressing Frictional Force**:
- The frictional force can be expressed in terms of the coefficient of static friction (μs):
\[ f = \mu_s N \]
- Substituting this into the horizontal force equation gives:
\[ N \sin \theta + \mu_s N \cos \theta = \frac{mv^2}{R} \]

4. **Substituting for Normal Force**:
- From the vertical force balance, we can express N:
\[ N = \frac{mg + \mu_s N \sin \theta}{\cos \theta} \]
- Rearranging gives:
\[ N \cos \theta = mg + \mu_s N \sin \theta \]

5. **Combining the Equations**:
- We can substitute the expression for N from the vertical balance into the horizontal balance:
\[ \frac{(mg + \mu_s N \sin \theta) \sin \theta}{\cos \theta} + \mu_s \frac{(mg + \mu_s N \sin \theta) \cos \theta}{\cos \theta} = \frac{mv^2}{R} \]

6. **Solving for Maximum Velocity**:
- After simplifying and rearranging the equations, we arrive at:
\[ v_{\text{max}}^2 = Rg \left( \tan \theta + \mu_s \right) \div \left( 1 - \mu_s \tan \theta \right) \]
- Taking the square root gives:
\[ v_{\text{max}} = \sqrt{Rg \left( \tan \theta + \mu_s \right) \div \left( 1 - \mu_s \tan \theta \right)} \]

### Final Result:
The maximum safe velocity \( v_{\text{max}} \) is given by:
\[ v_{\text{max}} = \sqrt{Rg \left( \tan \theta + \mu_s \right) \div \left( 1 - \mu_s \tan \theta \right)} \]