Consider a car moving on a straight road with a speed of `100m//s`. The distance at which car can be stopped is `[mu_k=0.5]`

Options

A. (a) 1000m
B. (b) 800m
C. (c) 400m
D. (d) 100m

Explanation

To solve the problem of determining the distance at which a car moving at a speed of 100 m/s can be stopped given a coefficient of kinetic friction (μ_k) of 0.5, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Initial speed of the car (u) = 100 m/s
- Coefficient of kinetic friction (μ_k) = 0.5
- Acceleration due to gravity (g) = 10 m/s² (assuming standard value)

2. **Calculate the Retardation (a)**:
The force of kinetic friction (F_k) acting on the car can be calculated using:
\[ F_k = \mu_k \cdot m \cdot g \]
The acceleration (retardation in this case) can be found using Newton's second law:
\[ F = m \cdot a \implies a = \frac{F_k}{m} = \mu_k \cdot g \]
Substituting the values:
\[ a = 0.5 \cdot 10 = 5 \text{ m/s}^2 \]
Since this is retardation, we take it as negative:
\[ a = -5 \text{ m/s}^2 \]

3. **Apply the Third Equation of Motion**:
We will use the third equation of motion to find the stopping distance (S):
\[ v^2 = u^2 + 2aS \]
Here, the final velocity (v) when the car stops is 0 m/s. Substituting the known values:
\[ 0 = (100)^2 + 2(-5)S \]
This simplifies to:
\[ 0 = 10000 - 10S \]

4. **Solve for S**:
Rearranging the equation to solve for S:
\[ 10S = 10000 \implies S = \frac{10000}{10} = 1000 \text{ meters} \]

5. **Conclusion**:
The distance at which the car can be stopped is **1000 meters**.