A car moving with a speed of `40 km//h` can be stopped by applying the brakes after at least 2 m. If the same car is moving with a speed of `80 km//h`, what is the minimum stopping distance?

Options

A. `8 m`
B. `2 m`
C. `4 m`
D. `6 m`

Explanation

To solve the problem of finding the minimum stopping distance of a car moving at 80 km/h, given that it can stop from 40 km/h in a distance of 2 meters, we can follow these steps:

### Step 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 40 km/h:
\[ U_1 = 40 \times \frac{5}{18} = \frac{200}{18} \approx 11.11 \, \text{m/s} \]
- For 80 km/h:
\[ U_2 = 80 \times \frac{5}{18} = \frac{400}{18} \approx 22.22 \, \text{m/s} \]

### Step 2: Use the equation of motion to find acceleration
- We know that the car stops from 40 km/h (11.11 m/s) in 2 meters. We can use the equation:
\[ V^2 = U^2 + 2aS \]
where:
- \( V = 0 \) (final velocity)
- \( U = U_1 \) (initial velocity)
- \( S = 2 \, \text{m} \) (stopping distance)
- \( a \) is the acceleration (deceleration in this case, so it will be negative).

- Plugging in the values:
\[ 0 = (11.11)^2 + 2a(2) \]
\[ 0 = 123.456 + 4a \]
\[ 4a = -123.456 \implies a = -30.864 \, \text{m/s}^2 \]

### Step 3: Calculate the stopping distance for 80 km/h
- Now we will use the same equation for the speed of 80 km/h (22.22 m/s):
\[ V^2 = U^2 + 2aS \]
where:
- \( V = 0 \)
- \( U = U_2 = 22.22 \, \text{m/s} \)
- \( a = -30.864 \, \text{m/s}^2 \)

- Plugging in the values:
\[ 0 = (22.22)^2 + 2(-30.864)S \]
\[ 0 = 493.7284 - 61.728S \]
\[ 61.728S = 493.7284 \implies S = \frac{493.7284}{61.728} \approx 8 \, \text{m} \]

### Conclusion
The minimum stopping distance when the car is moving at 80 km/h is approximately **8 meters**.

---