The maximum speed (in `ms^(-1)` ) with which a car driver can traverse on a horizontal unbanked curve of radius 80 m with coefficient of friction 0.5 without skidding is : (g = 10 m/`s^(2)`)

Options

A. 25 m/s
B. 20 m/s
C. 15 m/s
D. depends on mass of the car

Explanation

To find the maximum speed with which a car driver can traverse a horizontal unbanked curve without skidding, we can follow these steps:

### Step 1: Understand the forces acting on the car
When a car is moving in a circular path, the forces acting on it include:
- The gravitational force (weight) acting downward, which is \( mg \).
- The normal force acting upward, which balances the gravitational force.
- The frictional force that provides the necessary centripetal force for circular motion.

### Step 2: Identify the relevant equations
For a car moving in a circular path, the centripetal force required is given by:
\[ F_c = \frac{mv^2}{r} \]
where:
- \( m \) is the mass of the car,
- \( v \) is the speed of the car,
- \( r \) is the radius of the curve.

The frictional force can be expressed as:
\[ F_f = \mu N \]
where:
- \( \mu \) is the coefficient of friction,
- \( N \) is the normal force.

Since the road is horizontal and unbanked, the normal force \( N \) is equal to the gravitational force \( mg \):
\[ N = mg \]
Thus, the frictional force becomes:
\[ F_f = \mu mg \]

### Step 3: Set the centripetal force equal to the frictional force
For the car to navigate the turn without skidding, the frictional force must equal the centripetal force:
\[ \mu mg = \frac{mv^2}{r} \]

### Step 4: Cancel out the mass \( m \)
Since \( m \) appears on both sides of the equation, we can cancel it out:
\[ \mu g = \frac{v^2}{r} \]

### Step 5: Solve for \( v^2 \)
Rearranging the equation gives:
\[ v^2 = \mu g r \]

### Step 6: Substitute the known values
Given:
- \( \mu = 0.5 \)
- \( g = 10 \, \text{m/s}^2 \)
- \( r = 80 \, \text{m} \)

Substituting these values into the equation:
\[ v^2 = 0.5 \times 10 \times 80 \]
\[ v^2 = 400 \]

### Step 7: Calculate \( v \)
Taking the square root of both sides:
\[ v = \sqrt{400} = 20 \, \text{m/s} \]

### Conclusion
The maximum speed with which the car driver can traverse the curve without skidding is \( 20 \, \text{m/s} \).

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