Consider a car moving along a straight horizontal road with a speed of 72 km / h . If the coefficient of kinetic friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is `[g=10ms^(-2)]`

Options

A. 30 m
B. 40 m
C. 72 m
D. 20 m

Answer

To find the shortest distance in which the car can be stopped, we can follow these steps:

### Step 1: Convert the speed from km/h to m/s
The initial speed of the car is given as 72 km/h. We need to convert this to meters per second (m/s).

\[ \text{Speed in m/s} = \text{Speed in km/h} \times \frac{5}{18} \]

Substituting the value:

\[ \text{Speed in m/s} = 72 \times \frac{5}{18} = 20 \, \text{m/s} \]

### Step 2: Calculate the frictional force
The frictional force \( F \) can be calculated using the formula:

\[ F = \mu \cdot N \]

Where:
- \( \mu \) is the coefficient of kinetic friction (0.5)
- \( N \) is the normal force, which is equal to the weight of the car \( mg \) (where \( g = 10 \, \text{m/s}^2 \))

Thus, the frictional force becomes:

\[ F = \mu \cdot mg = 0.5 \cdot mg \]

### Step 3: Calculate the acceleration
The acceleration \( a \) can be found using Newton's second law:

\[ F = ma \implies a = \frac{F}{m} \]

Substituting the expression for frictional force:

\[ a = \frac{0.5 \cdot mg}{m} = 0.5g \]

Substituting \( g = 10 \, \text{m/s}^2 \):

\[ a = 0.5 \cdot 10 = 5 \, \text{m/s}^2 \]

Since the car is decelerating, we take acceleration as negative:

\[ a = -5 \, \text{m/s}^2 \]

### Step 4: Use the kinematic equation to find the stopping distance
We can use the kinematic equation:

\[ V^2 = U^2 + 2aS \]

Where:
- \( V \) is the final velocity (0 m/s, since the car stops)
- \( U \) is the initial velocity (20 m/s)
- \( a \) is the acceleration (-5 m/s²)
- \( S \) is the distance we want to find

Rearranging the equation to solve for \( S \):

\[ 0 = (20)^2 + 2(-5)S \]

\[ 0 = 400 - 10S \]

\[ 10S = 400 \]

\[ S = \frac{400}{10} = 40 \, \text{m} \]

### Final Answer
The shortest distance in which the car can be stopped is **40 meters**.

---