The brakes of a car moving at 20 m/s along a horizontal road are suddenly applied and it comes to rest after travelling some distance if the coefficient of friction between the tyres and the road is 0.90 and it is assumed that all four tyres behave identically, the shortest distance the car would travel before coming to a stop is

Options

A. 2.27 m
B. 11.35 m
C. 22.7 m
D. 4.54 m

Answer

To solve the problem of determining the shortest distance a car would travel before coming to a stop after the brakes are applied, we can follow these steps:

### Step 1: Identify the given values
- Initial velocity (u) = 20 m/s
- Coefficient of friction (μ) = 0.90
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Calculate the deceleration (retardation) due to friction
The frictional force (f_k) acting on the car can be expressed as:
\[ f_k = \mu \cdot N \]
Where N is the normal force. For a car on a horizontal road, the normal force (N) equals the weight of the car (mg). Thus, we have:
\[ f_k = \mu \cdot mg \]

Using Newton's second law, the net force acting on the car when brakes are applied is equal to the mass times the acceleration (deceleration in this case):
\[ ma = -f_k \]
Substituting for f_k, we get:
\[ ma = -\mu mg \]

Dividing both sides by m, we find:
\[ a = -\mu g \]
Substituting the values:
\[ a = -0.90 \cdot 9.8 \]
\[ a = -8.82 \, \text{m/s}^2 \]

### Step 3: Use the equation of motion to find the stopping distance
We can use the equation of motion:
\[ v^2 = u^2 + 2as \]
Where:
- v = final velocity = 0 m/s (the car comes to rest)
- u = initial velocity = 20 m/s
- a = -8.82 m/s² (deceleration)
- s = distance traveled before stopping

Rearranging the equation to solve for s:
\[ 0 = (20)^2 + 2(-8.82)s \]
\[ 0 = 400 - 17.64s \]
\[ 17.64s = 400 \]
\[ s = \frac{400}{17.64} \]
\[ s \approx 22.67 \, \text{m} \]

### Conclusion
The shortest distance the car would travel before coming to a stop is approximately **22.67 meters**.

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