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

Answer

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

### Step 1: Understand the relationship between speed, stopping distance, and retardation.
The stopping distance \( s \) of a vehicle can be expressed using the formula:
\[ s = \frac{u^2}{2a} \]
where \( u \) is the initial speed and \( a \) is the retardation (deceleration).

### Step 2: Calculate the retardation using the first case.
For the first case where the car is moving at 40 km/h and stops in 2 m:
- Convert the speed from km/h to m/s:
\[ u_1 = 40 \text{ km/h} = \frac{40 \times 1000}{3600} = \frac{40000}{3600} \approx 11.11 \text{ m/s} \]
- Using the stopping distance formula:
\[ s_1 = 2 \text{ m} \]
We can rearrange the formula to find \( a \):
\[ a = \frac{u_1^2}{2s_1} \]
Substituting the values:
\[ a = \frac{(11.11)^2}{2 \times 2} = \frac{123.4569}{4} \approx 30.8642 \text{ m/s}^2 \]

### Step 3: Calculate the stopping distance for the second case.
Now, for the second case where the car is moving at 80 km/h:
- Convert the speed from km/h to m/s:
\[ u_2 = 80 \text{ km/h} = \frac{80 \times 1000}{3600} = \frac{80000}{3600} \approx 22.22 \text{ m/s} \]

### Step 4: Use the stopping distance formula for the second case.
Using the same retardation \( a \) calculated from the first case:
\[ s_2 = \frac{u_2^2}{2a} \]
Substituting the values:
\[ s_2 = \frac{(22.22)^2}{2 \times 30.8642} \]
Calculating \( (22.22)^2 \):
\[ (22.22)^2 \approx 493.7284 \]
Now substituting this value:
\[ s_2 = \frac{493.7284}{61.7284} \approx 8 \text{ m} \]

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

---