A car is moving along a straight horizontal road with a speed `v_(0)` . If the coefficient of friction between the tyre 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))/(mug)`
C. `((v_(0))/(mug))^(2)`
D. `(v_(0))/(mu)`

Explanation

To find the shortest distance in which a car can be stopped when moving with an initial speed \( v_0 \) and having a coefficient of friction \( \mu \) between the tires and the road, we can follow these steps:

### Step 1: Identify the forces acting on the car
The main force acting to stop the car is the frictional force between the tires and the road. The maximum frictional force \( F_f \) can be expressed as:
\[ F_f = \mu N \]
where \( N \) is the normal force. For a car moving on a horizontal road, the normal force \( N \) equals the weight of the car \( mg \), where \( m \) is the mass of the car and \( g \) is the acceleration due to gravity. Thus,
\[ F_f = \mu mg \]

### Step 2: Relate frictional force to acceleration
According to Newton's second law, the net force acting on the car is equal to the mass of the car times its acceleration \( a \):
\[ F = ma \]
In this case, the frictional force is the only force acting in the direction opposite to the motion, so we can write:
\[ ma = -F_f \]
Substituting for \( F_f \):
\[ ma = -\mu mg \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ a = -\mu g \]
This indicates that the acceleration (or retardation in this case) is \( -\mu g \).

### Step 3: Use the equations of motion
We can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance:
\[ v^2 = u^2 + 2as \]
Here, \( v \) is the final velocity (which is 0 when the car stops), \( u \) is the initial velocity \( v_0 \), \( a \) is the acceleration \( -\mu g \), and \( s \) is the distance we want to find (the stopping distance).

Substituting the known values into the equation:
\[ 0 = v_0^2 + 2(-\mu g)s \]
This simplifies to:
\[ 0 = v_0^2 - 2\mu gs \]

### Step 4: Solve for the stopping distance \( s \)
Rearranging the equation to solve for \( s \):
\[ 2\mu gs = v_0^2 \]
\[ s = \frac{v_0^2}{2\mu g} \]

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