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 step by step, we can follow these steps:

### Step 1: Understand the relationship between distance and initial velocity
When a car applies brakes, it comes to a stop due to deceleration. The distance covered before stopping is related to the initial velocity of the car. The formula that relates the initial velocity (u), final velocity (v), acceleration (a), and distance (s) is given by:
\[ v^2 = u^2 + 2as \]
In our case, the final velocity \( v = 0 \) (the car stops), so the equation simplifies to:
\[ 0 = u^2 + 2as \]
This can be rearranged to:
\[ s = -\frac{u^2}{2a} \]
This shows that the stopping distance \( s \) is directly proportional to the square of the initial velocity \( u \).

### Step 2: Set up the known values
From the problem, we know:
- The initial velocity of the car is \( V \).
- The distance covered when the car stops at this velocity is \( s_1 = 20 \, \text{m} \).

### Step 3: Establish the relationship for the doubled velocity
If the velocity of the car is doubled, the new initial velocity becomes:
\[ u_2 = 2V \]
Using the relationship established earlier, we can express the new stopping distance \( s_2 \) in terms of \( s_1 \):
\[ s_2 \propto u^2 \]
Thus, we can write:
\[ \frac{s_2}{s_1} = \left(\frac{u_2}{u_1}\right)^2 \]
Substituting \( u_1 = V \) and \( u_2 = 2V \):
\[ \frac{s_2}{s_1} = \left(\frac{2V}{V}\right)^2 = 4 \]

### Step 4: Calculate the new distance
Now, we can find \( s_2 \):
\[ s_2 = 4 \times s_1 \]
Substituting \( s_1 = 20 \, \text{m} \):
\[ s_2 = 4 \times 20 \, \text{m} = 80 \, \text{m} \]

### Conclusion
The distance the car will cover after applying brakes when its velocity is doubled is:
\[ \boxed{80 \, \text{m}} \]

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