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

Options

A. `(V_(0)^(2))/(2mug)`
B. `(V_(0)^(2))/(mug)`
C. `((V_(0))/(mug))^(2)`
D. `(2V_(0)^(2))/(mug)`

Explanation

To solve the problem of finding the shortest distance in which a car can be stopped while moving with an initial speed \( v_0 \) and given the coefficient of friction \( \mu \), we can use the work-energy principle. Here's a step-by-step solution:

### Step 1: Identify the forces acting on the car
The forces acting on the car include:
- The gravitational force \( mg \) acting downwards.
- The normal force \( N \) acting upwards.
- The frictional force \( f \) acting opposite to the direction of motion.

Since the car is on a horizontal road and not accelerating vertically, the normal force \( N \) is equal to the gravitational force:
\[ N = mg \]

### Step 2: Determine the frictional force
The frictional force, which is what brings the car to a stop, is given by:
\[ f = \mu N = \mu mg \]

### Step 3: Apply the work-energy theorem
According to the work-energy theorem, the work done by the frictional force is equal to the change in kinetic energy of the car. The work done by the frictional force over a distance \( s \) is:
\[ W = -f \cdot s = -\mu mg \cdot s \]
The negative sign indicates that the work done by friction is in the opposite direction of the displacement.

The initial kinetic energy of the car is:
\[ KE_i = \frac{1}{2} mv_0^2 \]
The final kinetic energy when the car comes to a stop is:
\[ KE_f = 0 \]
Thus, the change in kinetic energy is:
\[ \Delta KE = KE_f - KE_i = 0 - \frac{1}{2} mv_0^2 = -\frac{1}{2} mv_0^2 \]

### Step 4: Set up the equation
According to the work-energy principle:
\[ -\mu mg s = -\frac{1}{2} mv_0^2 \]
We can cancel the mass \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \mu g s = \frac{1}{2} v_0^2 \]

### Step 5: Solve for the stopping distance \( s \)
Rearranging the equation to solve for \( s \):
\[ s = \frac{v_0^2}{2 \mu g} \]

### Final Result
Thus, the shortest distance in which the car can be stopped is:
\[ s = \frac{v_0^2}{2 \mu g} \]