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))/(mug)`
C. `((v_(0))/(mug))^(2)`
D. `(v_(0))/(mu)`

Answer

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

### Step-by-Step Solution:

1. **Identify the forces acting on the car**:
When the car is moving, the only horizontal force acting to stop the car is the frictional force. This force is given by:
\[ F = \mu N \]
where \( N \) is the normal force. For a car on a horizontal road, \( N = mg \), where \( m \) is the mass of the car and \( g \) is the acceleration due to gravity. Thus, the frictional force becomes:
\[ F = \mu mg \]

2. **Determine the acceleration**:
The acceleration \( a \) of the car can be found using Newton's second law, \( F = ma \). Rearranging gives:
\[ a = \frac{F}{m} = \frac{\mu mg}{m} = \mu g \]
Since this force acts in the opposite direction to the motion, the acceleration will be negative:
\[ a = -\mu g \]

3. **Use the kinematic equation**:
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 (0, since 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. Substituting these values in gives:
\[ 0 = v_0^2 + 2(-\mu g)s \]

4. **Rearranging the equation**:
Rearranging the equation to solve for \( s \):
\[ v_0^2 = 2\mu gs \]
\[ s = \frac{v_0^2}{2\mu g} \]

5. **Final result**:
Thus, the shortest distance \( s \) in which the car can be stopped is:
\[ s = \frac{v_0^2}{2\mu g} \]