Question

A car is moving along a straight horizontal road with a speed \(v_0\). If the coefficient of friction between the tyres and the road is \(\mu\), the shortest distance in which the car can be stopped is

• A. \(\frac{v_0^2}{\mu g}\)
• B. \(\left(\frac{v_0}{\mu g}\right)^2\)
• C. \(\frac{v_0^2}{\mu g}\)
• D. \(\frac{v_0^2}{2\mu g}\)

Hint:

Stopping distance is independent of mass of the car.

Answer 1

The Correct Option is D

To find the shortest distance in which the car can be stopped, we need to apply the basic principles of kinematics and friction. The problem involves a car moving with an initial velocity \( v_0 \) and is brought to rest using frictional force. The shortest stopping distance \( s \) can be found using the equation derived from the work-energy principle.

1. **Work-Energy Principle**:

The work done by the frictional force brings the car to rest. The work done by friction is equal to the initial kinetic energy of the car.

• Initial Kinetic Energy, \( KE_i = \frac{1}{2} m v_0^2 \)
• Work done by friction, \( W = \text{friction force} \times \text{distance} = \mu mg \cdot s \)

Since the car comes to rest, the work done by the friction (negative work) equals the loss in kinetic energy:

\[-\mu mg \cdot s = -\frac{1}{2} m v_0^2\]

2. **Solve for the Stopping Distance**:

Cancel the mass \( m \) from both sides of the equation, which gives:

\[\mu g \cdot s = \frac{1}{2} v_0^2\]

Rearranging for \( s \), we get:

\[s = \frac{v_0^2}{2\mu g}\]

This formula expresses the shortest stopping distance in terms of initial velocity \( v_0 \), coefficient of friction \( \mu \), and acceleration due to gravity \( g \).

Therefore, the correct answer is:

\(\frac{v_0^2}{2\mu g}\)

3. **Explanation of Options**:

• \(\frac{v_0^2}{\mu g}\): This would be the stopping distance if the frictional force is just balancing some fraction of kinetic energy, not considering complete stop.
• \(\left(\frac{v_0}{\mu g}\right)^2\) and \(\frac{v_0^2}{\mu g}\): Incorrect dimensional analysis and don't provide correct stopping distance.
• \(\frac{v_0^2}{2\mu g}\): Correct answer derived logically.

Therefore, the shortest stopping distance, when the car is subjected to friction, is accurately represented by: \(\frac{v_0^2}{2\mu g}\).