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

Explanation

To solve the problem, we will use the equations of motion and the concept of braking force.

### Given:
1. Initial speed of the car, \( u_1 = 30 \) km/h
2. Stopping distance at this speed, \( s_1 = 8 \) m
3. Initial speed of the car when it is going to be brought to a halt again, \( u_2 = 60 \) km/h

### Step 1: Convert speeds from km/h to m/s
To convert km/h to m/s, we use the conversion factor \( \frac{5}{18} \).

\[ u_1 = 30 \times \frac{5}{18} = \frac{150}{18} = 8.33 \text{ m/s} \]

\[ u_2 = 60 \times \frac{5}{18} = \frac{300}{18} = 16.67 \text{ m/s} \]

### Step 2: Use the equation of motion to find acceleration
We can use the equation of motion:

\[ v^2 = u^2 + 2as \]

Where:
- \( v \) = final velocity (0 m/s when the car stops)
- \( u \) = initial velocity
- \( a \) = acceleration (which will be negative since it is deceleration)
- \( s \) = stopping distance

For the first case (when \( u = u_1 \)):

\[ 0 = (8.33)^2 + 2a(8) \]

\[ 0 = 69.39 + 16a \]

\[ 16a = -69.39 \]

\[ a = -\frac{69.39}{16} \approx -4.35 \text{ m/s}^2 \]

### Step 3: Use the same acceleration to find stopping distance for the second speed
Now, we will use the same acceleration to find the stopping distance when the car is traveling at \( u_2 \).

Using the same equation of motion:

\[ 0 = (16.67)^2 + 2(-4.35)s_2 \]

\[ 0 = 278.89 - 8.7s_2 \]

\[ 8.7s_2 = 278.89 \]

\[ s_2 = \frac{278.89}{8.7} \approx 32.0 \text{ m} \]

### Conclusion
The car traveling at 60 km/h can be brought to a halt in approximately **32 meters**.

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