A car moving with a velocity of 40 km/h can be stopped by brekes after travelling at least 5 m. If same car moves at a speed of 80 km/h the minimum stopping distance is

Options

A. 10 m
B. 5 m
C. 15 m
D. 20 m

Answer

To solve the problem, we will use the relationship between the initial velocity, stopping distance, and acceleration (or deceleration) of the car. We can derive the stopping distance for the car moving at a higher speed based on the stopping distance at the lower speed.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial speed \( u_1 = 40 \) km/h
- Stopping distance \( s_1 = 5 \) m
- New speed \( u_2 = 80 \) km/h
- We need to find the new stopping distance \( s_2 \).

2. **Convert Speeds from km/h to m/s:**
- To convert km/h to m/s, we use the conversion factor \( \frac{5}{18} \).
- \( u_1 = 40 \times \frac{5}{18} = \frac{200}{18} \approx 11.11 \) m/s
- \( u_2 = 80 \times \frac{5}{18} = \frac{400}{18} \approx 22.22 \) m/s

3. **Use the Equation of Motion:**
- The equation of motion we will use is:
\[ v^2 = u^2 + 2as \]
- For both cases, the final velocity \( v = 0 \) (the car stops), so we can rearrange the equation to:
\[ 0 = u^2 + 2as \implies a = -\frac{u^2}{2s} \]

4. **Set Up the Equations for Both Cases:**
- For the first case:
\[ a = -\frac{u_1^2}{2s_1} \]
- For the second case:
\[ a = -\frac{u_2^2}{2s_2} \]

5. **Equate the Two Expressions for Acceleration:**
- Since the car is the same in both cases, the acceleration (deceleration) is the same:
\[ -\frac{u_1^2}{2s_1} = -\frac{u_2^2}{2s_2} \]
- This simplifies to:
\[ \frac{u_1^2}{s_1} = \frac{u_2^2}{s_2} \]

6. **Rearranging to Find \( s_2 \):**
- Rearranging gives us:
\[ s_2 = s_1 \cdot \frac{u_2^2}{u_1^2} \]

7. **Substituting Known Values:**
- Substitute \( s_1 = 5 \) m, \( u_1 = 11.11 \) m/s, and \( u_2 = 22.22 \) m/s:
\[ s_2 = 5 \cdot \left(\frac{22.22^2}{11.11^2}\right) \]

8. **Calculating the Ratio:**
- The ratio \( \frac{u_2}{u_1} = \frac{22.22}{11.11} = 2 \).
- Therefore, \( \left(\frac{u_2}{u_1}\right)^2 = 2^2 = 4 \).

9. **Final Calculation:**
- Thus,
\[ s_2 = 5 \cdot 4 = 20 \text{ m} \]

### Conclusion:
The minimum stopping distance \( s_2 \) when the car is moving at 80 km/h is **20 meters**.