A car is moving on a straight horizontal road with a speed v. When brakes are applied to give a constant retardation a, the car is stopped in a shortest distance S. If the car was moving on the same road with a speed 3v and the same retardation a is applied, the shortest distance in which the car is stopped will be

Options

A. 3S
B. 6S
C. 9S
D. 27S

Explanation

To solve the problem, we need to analyze the motion of the car under constant retardation. We will use the third equation of motion, which relates initial velocity, final velocity, acceleration, and distance traveled.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Initial speed of the car, \( v \)
- Final speed of the car when it stops, \( 0 \)
- Constant retardation (deceleration), \( a \)
- Distance covered to stop from speed \( v \), \( S \)

2. **Use the Third Equation of Motion:**
The third equation of motion states:
\[ v_f^2 = v_i^2 + 2a s \]
Where:
- \( v_f \) = final velocity (0 when the car stops)
- \( v_i \) = initial velocity
- \( a \) = acceleration (negative for retardation)
- \( s \) = distance

For the first scenario (speed \( v \)):
\[ 0 = v^2 - 2aS \]
Rearranging gives:
\[ v^2 = 2aS \quad \text{(1)} \]

3. **Consider the Second Scenario:**
Now, the car is moving with an initial speed of \( 3v \) and the same retardation \( a \). We need to find the new stopping distance \( S_1 \).

4. **Apply the Third Equation Again:**
For the second scenario (speed \( 3v \)):
\[ 0 = (3v)^2 - 2aS_1 \]
Rearranging gives:
\[ (3v)^2 = 2aS_1 \]
Simplifying this:
\[ 9v^2 = 2aS_1 \quad \text{(2)} \]

5. **Relate the Two Equations:**
From equation (1), we have:
\[ v^2 = 2aS \]
We can substitute \( 2a \) from equation (1) into equation (2):
\[ 9v^2 = 9 \cdot (2aS) \]
Therefore:
\[ S_1 = \frac{9v^2}{2a} \]

6. **Express \( S_1 \) in Terms of \( S \):**
Since \( v^2 = 2aS \), we can replace \( v^2 \) in the expression for \( S_1 \):
\[ S_1 = 9 \cdot S \]

### Final Result:
Thus, the shortest distance in which the car is stopped when moving at a speed of \( 3v \) is:
\[ S_1 = 9S \]