A car travelling at a speed of 30 km / hour is brought to a halt in 8 m by applying brakes. If the same car is travelling at 60 km / hour , it can be brought to a halt with the same braking force in

Options

A. 8 m
B. 16 m
C. 24 m
D. 32 m

Answer

To solve the problem, we need to determine how far a car traveling at 60 km/h can be brought to a halt using the same braking force that was used to stop it from 30 km/h over a distance of 8 meters. We will use the work-energy principle, which states that the work done by the braking force is equal to the change in kinetic energy of the car.

### Step-by-Step Solution:

1. **Convert Speeds from km/h to m/s**:
- Speed at 30 km/h:
\[ v_1 = 30 \text{ km/h} = \frac{30 \times 1000}{3600} = \frac{30000}{3600} \approx 8.33 \text{ m/s} \]
- Speed at 60 km/h:
\[ v_2 = 60 \text{ km/h} = \frac{60 \times 1000}{3600} = \frac{60000}{3600} \approx 16.67 \text{ m/s} \]

2. **Calculate Initial Kinetic Energy at 30 km/h**:
- The kinetic energy (KE) formula is:
\[ KE = \frac{1}{2} mv^2 \]
- Let \( m \) be the mass of the car. The initial kinetic energy when traveling at 30 km/h is:
\[ KE_1 = \frac{1}{2} m (8.33)^2 \]

3. **Calculate Initial Kinetic Energy at 60 km/h**:
- The initial kinetic energy when traveling at 60 km/h is:
\[ KE_2 = \frac{1}{2} m (16.67)^2 \]

4. **Set Up the Work-Energy Principle**:
- The work done by the braking force is equal to the change in kinetic energy:
\[ \text{Work} = KE_1 - 0 = \frac{1}{2} m (8.33)^2 \]
\[ \text{Work} = KE_2 - 0 = \frac{1}{2} m (16.67)^2 \]

5. **Relate Work Done to Distance**:
- The work done can also be expressed as:
\[ \text{Work} = F \cdot d \]
- For the first scenario (30 km/h):
\[ F \cdot 8 = \frac{1}{2} m (8.33)^2 \]
- For the second scenario (60 km/h):
\[ F \cdot d_2 = \frac{1}{2} m (16.67)^2 \]

6. **Using the same braking force**:
- Since the braking force \( F \) is the same in both cases, we can set up the ratio of distances:
\[ \frac{d_1}{d_2} = \frac{(v_1)^2}{(v_2)^2} \]
- Plugging in the values:
\[ \frac{8}{d_2} = \frac{(8.33)^2}{(16.67)^2} \]

7. **Calculate \( d_2 \)**:
- Rearranging gives:
\[ d_2 = 8 \cdot \left(\frac{(16.67)^2}{(8.33)^2}\right) \]
- Calculate:
\[ d_2 = 8 \cdot \left(\frac{(16.67)^2}{(8.33)^2}\right) = 8 \cdot \left(\frac{278.0889}{69.3889}\right) \approx 32 \text{ meters} \]

### Final Answer:
The car traveling at 60 km/h can be brought to a halt with the same braking force in approximately **32 meters**.