A car, moving with a speed of `50 km//hr`, can be stopped by brakes after at least `6 m`. If the same car is moving at a speed of `100 km//hr`, the minimum stopping distance is

Options

A. 6 m
B. 2 m
C. 18 m
D. 24 m

Explanation

To solve the problem, we will use the work-energy theorem, which states that the work done by the stopping force is equal to the change in kinetic energy of the car.

### Step-by-Step Solution:

1. **Identify the given data:**
- Initial speed of the car, \( v_1 = 50 \, \text{km/hr} \)
- Stopping distance at this speed, \( x_1 = 6 \, \text{m} \)
- New speed of the car, \( v_2 = 100 \, \text{km/hr} \)

2. **Convert speeds from km/hr to m/s:**
- To convert km/hr to m/s, use the conversion factor \( \frac{5}{18} \).
- For \( v_1 \):
\[ v_1 = 50 \times \frac{5}{18} = \frac{250}{18} \approx 13.89 \, \text{m/s} \]
- For \( v_2 \):
\[ v_2 = 100 \times \frac{5}{18} = \frac{500}{18} \approx 27.78 \, \text{m/s} \]

3. **Use the work-energy theorem:**
- The work done by the stopping force is equal to the change in kinetic energy.
- The formula for kinetic energy is \( KE = \frac{1}{2} mv^2 \).
- The work done by the stopping force can be expressed as:
\[ W = F_s \cdot x \]
- Since the stopping force \( F_s \) is constant, we can set up the following relationship for both speeds:
\[ F_s \cdot x_1 = \frac{1}{2} m v_1^2 \quad \text{(1)} \]
\[ F_s \cdot x_2 = \frac{1}{2} m v_2^2 \quad \text{(2)} \]

4. **Divide equation (1) by equation (2):**
- This gives:
\[ \frac{F_s \cdot x_1}{F_s \cdot x_2} = \frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} \]
- The mass \( m \) and the factor \( \frac{1}{2} \) cancel out:
\[ \frac{x_1}{x_2} = \frac{v_1^2}{v_2^2} \]

5. **Substitute the known values:**
- We know \( x_1 = 6 \, \text{m} \), \( v_1 = 13.89 \, \text{m/s} \), and \( v_2 = 27.78 \, \text{m/s} \):
\[ \frac{6}{x_2} = \left(\frac{13.89}{27.78}\right)^2 \]
- Simplifying the right side:
\[ \frac{6}{x_2} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

6. **Solve for \( x_2 \):**
- Cross-multiplying gives:
\[ 6 \cdot 4 = x_2 \]
\[ x_2 = 24 \, \text{m} \]

### Final Answer:
The minimum stopping distance when the car is moving at a speed of \( 100 \, \text{km/hr} \) is \( 24 \, \text{m} \).

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