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. (a)25 m/s
B. (b)20 m/s
C. (c)15 m/s
D. (d)30 m/s

Explanation

To find the maximum speed with which a car can traverse a horizontal unbanked curve without skidding, we can use the relationship between centripetal force and frictional force. Here’s a step-by-step solution:

### Step 1: Understand the forces involved
For a car moving in a circular path, the centripetal force required to keep the car moving in that path must be provided by the frictional force between the tires and the road.

### Step 2: Write the equation for centripetal force
The centripetal force (F_c) required to keep the car moving in a circle 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.

### Step 3: Write the equation for frictional force
The maximum frictional force (F_f) that can act on the car is given by:
\[ F_f = \mu mg \]
where:
- \( \mu \) is the coefficient of friction,
- \( g \) is the acceleration due to gravity.

### Step 4: Set the forces equal
For the car to not skid, the centripetal force must equal the maximum frictional force:
\[ \frac{mv^2}{r} = \mu mg \]

### Step 5: Cancel out the mass
Since the mass \( m \) appears on both sides of the equation, we can cancel it out:
\[ \frac{v^2}{r} = \mu g \]

### Step 6: Rearrange the equation to solve for speed
Rearranging the equation gives us:
\[ v^2 = \mu g r \]
Taking the square root of both sides, we find:
\[ v = \sqrt{\mu g r} \]

### Step 7: Substitute the known values
Now we can substitute the values given in the problem:
- \( \mu = 0.5 \)
- \( g = 10 \, \text{m/s}^2 \)
- \( r = 80 \, \text{m} \)

Substituting these values into the equation:
\[ v = \sqrt{0.5 \times 10 \times 80} \]

### Step 8: Calculate the value
Calculating the expression inside the square root:
\[ v = \sqrt{0.5 \times 10 \times 80} = \sqrt{400} = 20 \, \text{m/s} \]

### Final Answer
The maximum speed with which the car can traverse the curve without skidding is:
\[ v = 20 \, \text{m/s} \]