The minimum velocity (in ms^(-1))` with which a car driver must traverse a flat curve of radius 150m and coefficient of friction 0.6 to avoid skidding is

Options

A. 60
B. 30
C. 15
D. 25

Answer

To find the minimum velocity with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding, we can follow these steps:

### Step 1: Understand the Forces Involved
When a car is moving in a circular path, it experiences a centripetal force that keeps it in that path. This centripetal force is provided by the frictional force between the tires and the road.

### Step 2: Identify the Relevant Equations
The centripetal force required to keep the car moving in a circle is given by:
\[ F_c = \frac{mv^2}{r} \]
where:
- \( m \) = mass of the car
- \( v \) = velocity of the car
- \( r \) = radius of the curve

The frictional force that prevents skidding is given by:
\[ F_f = \mu N \]
where:
- \( \mu \) = coefficient of friction
- \( N \) = normal force

On a flat surface, the normal force \( N \) is equal to the weight of the car, which is \( mg \):
\[ N = mg \]

### Step 3: Set the Forces Equal
For the car to avoid skidding, the frictional force must be equal to the centripetal force:
\[ \mu mg = \frac{mv^2}{r} \]

### Step 4: Simplify the Equation
We can cancel the mass \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \mu g = \frac{v^2}{r} \]

Rearranging gives us:
\[ v^2 = \mu g r \]

### Step 5: Substitute Known Values
Now we can substitute the known values into the equation:
- \( \mu = 0.6 \)
- \( g = 10 \, \text{m/s}^2 \) (approximate value of acceleration due to gravity)
- \( r = 150 \, \text{m} \)

So,
\[ v^2 = 0.6 \times 10 \times 150 \]

### Step 6: Calculate
Calculating the right side:
\[ v^2 = 0.6 \times 10 \times 150 = 900 \]

Now, taking the square root to find \( v \):
\[ v = \sqrt{900} = 30 \, \text{m/s} \]

### Conclusion
The minimum velocity with which the car driver must traverse the curve to avoid skidding is:
\[ v = 30 \, \text{m/s} \]