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

Options

A. `12 m `
B. `18 m`
C. `24 m`
D. ` 6 m`

Explanation

To solve the problem, we need to find the minimum stopping distance of a car moving at a speed of 100 km/h, given that a car moving at 50 km/h can stop in 6 meters. We will use the equations of motion to derive the stopping distance.

### Step-by-Step Solution:

1. **Convert Speeds from km/h to m/s**:
- The formula to convert km/h to m/s is:
\[ \text{Speed in m/s} = \text{Speed in km/h} \times \frac{5}{18} \]
- For 50 km/h:
\[ u_1 = 50 \times \frac{5}{18} = \frac{250}{18} \approx 13.89 \, \text{m/s} \]
- For 100 km/h:
\[ u_2 = 100 \times \frac{5}{18} = \frac{500}{18} \approx 27.78 \, \text{m/s} \]

2. **Use the Equation of Motion**:
- The equation we will use is:
\[ v^2 = u^2 + 2as \]
- Where:
- \(v\) = final velocity (0 m/s when the car stops)
- \(u\) = initial velocity
- \(a\) = acceleration (deceleration in this case)
- \(s\) = stopping distance

3. **Calculate Deceleration (a)**:
- From the first scenario (50 km/h):
- We know \(s_1 = 6 \, \text{m}\) and \(u_1 \approx 13.89 \, \text{m/s}\).
- Rearranging the equation:
\[ 0 = (13.89)^2 + 2a(6) \]
\[ 0 = 193.61 + 12a \]
\[ 12a = -193.61 \quad \Rightarrow \quad a \approx -16.13 \, \text{m/s}^2 \]

4. **Calculate Stopping Distance for 100 km/h**:
- Now we use the calculated deceleration to find the stopping distance when the car is traveling at 100 km/h:
- Using \(u_2 \approx 27.78 \, \text{m/s}\):
\[ 0 = (27.78)^2 + 2(-16.13)s_2 \]
\[ 0 = 771.84 - 32.26s_2 \]
\[ 32.26s_2 = 771.84 \quad \Rightarrow \quad s_2 = \frac{771.84}{32.26} \approx 23.95 \, \text{m} \]

5. **Final Answer**:
- Rounding off, the minimum stopping distance for the car moving at 100 km/h is approximately **24 meters**.

### Summary:
- The minimum stopping distance for a car moving at 100 km/h is **24 meters**.