A car is moving with velocity V. If stop after applying break at a distance of 20 m. If velocity of car is doubled, then how much distance it will cover (travel) after applying break :

Options

A. 40 m
B. 80 m
C. 160 m
D. 320 m

Explanation

To solve the problem, we can use the equations of motion. The key points to consider are the initial velocity of the car, the distance it travels before stopping, and how the distance changes when the initial velocity is doubled.

### Step-by-step Solution:

1. **Understanding the Variables**:
- Let the initial velocity of the car be \( V \).
- The distance covered before stopping when the car is moving with velocity \( V \) is given as \( S_1 = 20 \, \text{m} \).
- When the velocity is doubled, the new initial velocity becomes \( U = 2V \).

2. **Using the Equation of Motion**:
- We can use the third equation of motion, which states:
\[ V^2 = U^2 + 2aS \]
- In this case, when the car stops, the final velocity \( V = 0 \). Thus, we can rewrite the equation as:
\[ 0 = U^2 + 2aS \]
- Rearranging gives:
\[ S = -\frac{U^2}{2a} \]
- Here, \( S \) is the stopping distance, \( U \) is the initial velocity, and \( a \) is the deceleration.

3. **Finding the Relationship**:
- From the equation, we can see that the stopping distance \( S \) is proportional to the square of the initial velocity \( U^2 \):
\[ S \propto U^2 \]

4. **Setting Up Ratios**:
- Let \( S_2 \) be the stopping distance when the velocity is doubled:
\[ \frac{S_1}{S_2} = \frac{U_1^2}{U_2^2} \]
- Substituting \( U_1 = V \) and \( U_2 = 2V \):
\[ \frac{S_1}{S_2} = \frac{V^2}{(2V)^2} = \frac{V^2}{4V^2} = \frac{1}{4} \]

5. **Calculating the New Distance**:
- We know \( S_1 = 20 \, \text{m} \):
\[ \frac{20}{S_2} = \frac{1}{4} \]
- Cross-multiplying gives:
\[ S_2 = 20 \times 4 = 80 \, \text{m} \]

### Final Answer:
The distance the car will cover after applying the brakes when its velocity is doubled is **80 meters**.