An automobile travelling with a speed `60 km//h` , can brake to stop within a distance of `20 m` . If the car is going twice as fast i. e. , `120 km//h`, the stopping distance will be

Options

A. `60 m`
B. `40 m`
C. `20 m`
D. `80 m`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the speed from km/h to m/s
The initial speed \( u \) when the car is traveling at \( 60 \, \text{km/h} \) needs to be converted to meters per second (m/s):
\[ u = 60 \, \text{km/h} = 60 \times \frac{1000 \, \text{m}}{3600 \, \text{s}} = \frac{60 \times 1000}{3600} = 16.67 \, \text{m/s} \]

### Step 2: Use the stopping distance to find acceleration
We know that the stopping distance \( s \) is \( 20 \, \text{m} \) and the final velocity \( v \) is \( 0 \, \text{m/s} \). We can use the equation of motion:
\[ v^2 = u^2 + 2as \]
Substituting the known values:
\[ 0 = (16.67)^2 + 2a(20) \]
Rearranging gives:
\[ 2a(20) = -(16.67)^2 \]
\[ a = -\frac{(16.67)^2}{40} \]

### Step 3: Calculate the acceleration
Calculating \( a \):
\[ a = -\frac{(16.67)^2}{40} = -\frac{278.89}{40} \approx -6.97 \, \text{m/s}^2 \]

### Step 4: Calculate the stopping distance for the new speed
Now, we consider the case when the speed is \( 120 \, \text{km/h} \):
\[ u = 120 \, \text{km/h} = 120 \times \frac{1000 \, \text{m}}{3600 \, \text{s}} = \frac{120 \times 1000}{3600} = 33.33 \, \text{m/s} \]

Using the same equation of motion:
\[ v^2 = u^2 + 2as \]
Substituting the known values:
\[ 0 = (33.33)^2 + 2(-6.97)s \]
Rearranging gives:
\[ 2(-6.97)s = -(33.33)^2 \]
\[ s = \frac{(33.33)^2}{2 \times 6.97} \]

### Step 5: Calculate the stopping distance
Calculating \( s \):
\[ s = \frac{1111.09}{13.94} \approx 79.8 \, \text{m} \]

### Conclusion
The stopping distance when the car is traveling at \( 120 \, \text{km/h} \) is approximately \( 80 \, \text{m} \).

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