A car travelling at a speed of 30 km per hour is brought to rest in 8 metres by applying brakes. If the same car is travelling at 60 km per hour, it can be brought tor est with the same braking force at a distance of

Options

A. 16 m
B. 24 m
C. 32 m
D. 36 m

Answer

To solve the problem step by step, we will use the concepts of kinetic energy and the relationship between distance and speed when braking.

### Step 1: Understand the given information
- Initial speed \( u_1 = 30 \) km/h
- Final speed \( v = 0 \) (the car comes to rest)
- Distance \( d_1 = 8 \) m (the distance in which the car stops at 30 km/h)
- New speed \( u_2 = 60 \) km/h (the speed at which we need to find the stopping distance)

### Step 2: Convert speeds from km/h to m/s
To use the formulas correctly, we need to convert the speeds from kilometers per hour to meters per second:
- \( u_1 = 30 \) km/h = \( \frac{30 \times 1000}{3600} = 8.33 \) m/s
- \( u_2 = 60 \) km/h = \( \frac{60 \times 1000}{3600} = 16.67 \) m/s

### Step 3: Use the work-energy principle
The work done by the brakes is equal to the change in kinetic energy. The work done can be expressed as:
\[ W = F \cdot d \]
Where \( F \) is the braking force and \( d \) is the distance.

The change in kinetic energy is given by:
\[ \Delta KE = \frac{1}{2} m u^2 - \frac{1}{2} m v^2 \]
Since the car comes to rest, \( v = 0 \), so:
\[ \Delta KE = \frac{1}{2} m u^2 \]

### Step 4: Establish the relationship between distances and speeds
From the work-energy principle, we can say:
\[ F \cdot d_1 = \frac{1}{2} m u_1^2 \]
\[ F \cdot d_2 = \frac{1}{2} m u_2^2 \]

Since the braking force \( F \) and mass \( m \) are constant, we can set up the ratio:
\[ \frac{d_1}{d_2} = \frac{u_1^2}{u_2^2} \]

### Step 5: Substitute the known values
Substituting the known values into the equation:
\[ \frac{8}{d_2} = \frac{(8.33)^2}{(16.67)^2} \]

Calculating the squares:
- \( (8.33)^2 \approx 69.39 \)
- \( (16.67)^2 \approx 278.09 \)

Now substituting back:
\[ \frac{8}{d_2} = \frac{69.39}{278.09} \]

### Step 6: Solve for \( d_2 \)
Cross-multiplying gives:
\[ 8 \cdot 278.09 = 69.39 \cdot d_2 \]
\[ d_2 = \frac{8 \cdot 278.09}{69.39} \approx 32 \text{ m} \]

### Final Answer
The distance \( d_2 \) at which the car can be brought to rest when traveling at 60 km/h is approximately **32 meters**.

---