A car running with a velocity `72 kmph` on a level road is stoped after travelling a distance of `30m` after disengaging its engine `(g =10m^(-2))` The coefficient of friction between the road and the tyres is .

Options

A. `0.33`
B. `4.5`
C. `0.67`
D. `0.8`

Explanation

To solve the problem, we need to find the coefficient of friction between the road and the tires of the car. We will use the concepts of Newton's laws of motion and the equations of motion.

### Step-by-Step Solution:

1. **Convert the velocity from km/h to m/s:**
\[ \text{Velocity} (v) = 72 \text{ km/h} = \frac{72 \times 1000}{3600} \text{ m/s} = 20 \text{ m/s} \]

2. **Use the equation of motion to find the acceleration:**
We know that the car stops after traveling a distance of \( s = 30 \) m. The final velocity \( v_f = 0 \) m/s (since the car stops). We can use the equation:
\[ v_f^2 = v^2 + 2as \]
Rearranging it to find acceleration \( a \):
\[ 0 = (20)^2 + 2a(30) \]
\[ 0 = 400 + 60a \]
\[ 60a = -400 \]
\[ a = -\frac{400}{60} = -\frac{20}{3} \approx -6.67 \text{ m/s}^2 \]

3. **Relate the acceleration to the frictional force:**
The frictional force \( f \) is what brings the car to a stop. According to Newton's second law:
\[ f = ma \]
where \( m \) is the mass of the car and \( a \) is the acceleration. The frictional force can also be expressed as:
\[ f = \mu mg \]
where \( \mu \) is the coefficient of friction and \( g = 10 \text{ m/s}^2 \).

4. **Set the two expressions for frictional force equal to each other:**
\[ ma = \mu mg \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ a = \mu g \]

5. **Substituting the values to find \( \mu \):**
\[ -\frac{20}{3} = \mu \cdot 10 \]
\[ \mu = -\frac{20}{3 \times 10} = -\frac{2}{3} \]
Since the coefficient of friction cannot be negative, we take the absolute value:
\[ \mu = \frac{2}{3} \approx 0.67 \]

### Final Answer:
The coefficient of friction between the road and the tires is approximately \( \mu = 0.67 \).