An unloaded car moving with velocity u on a road can be stopped in a distance S. If passengers add `40%` to its weight and braking force remains the same, the stopping distance at velocity u is now

Options

A. 1.4 S
B. S
C. `(1.4)^(2)S`
D. `1//1.4 S`

Explanation

To solve the problem, we need to analyze the stopping distance of the car under two different conditions: first when it is unloaded, and second when it has additional weight from passengers.

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- Let the weight of the unloaded car be \( W \).
- The car stops in a distance \( S \) when it is moving with velocity \( u \).
- The braking force is constant and denoted as \( F_b \).

2. **Apply Newton's second law**:
- The net force acting on the car when braking is the braking force, which acts in the opposite direction of motion.
- According to Newton's second law, we have:
\[ -F_b = m \cdot a \]
- Here, \( m \) is the mass of the car, and \( a \) is the deceleration.

3. **Relate deceleration to stopping distance**:
- Using the kinematic equation:
\[ v^2 = u^2 + 2as \]
- When the car stops, \( v = 0 \), so:
\[ 0 = u^2 + 2(-a)S \]
- Rearranging gives:
\[ u^2 = 2aS \]
- Substituting \( a = -\frac{F_b}{m} \):
\[ u^2 = 2 \left(-\frac{F_b}{m}\right) S \]
- Therefore, we can express \( S \) as:
\[ S = \frac{m u^2}{2 F_b} \]

4. **Consider the second scenario with added weight**:
- When passengers add 40% to the weight of the car, the new weight becomes:
\[ W' = W + 0.4W = 1.4W \]
- The new mass \( m' \) of the car is:
\[ m' = 1.4m \]

5. **Determine the new deceleration**:
- The braking force remains the same, so applying Newton's second law again:
\[ -F_b = m' \cdot a' \]
- This gives:
\[ a' = -\frac{F_b}{m'} \]
- Substituting \( m' \):
\[ a' = -\frac{F_b}{1.4m} \]

6. **Relate new deceleration to new stopping distance**:
- Using the same kinematic equation:
\[ 0 = u^2 + 2a'S' \]
- Rearranging gives:
\[ u^2 = -2a'S' \]
- Substituting \( a' \):
\[ u^2 = 2 \left(-\frac{F_b}{1.4m}\right) S' \]
- Rearranging gives:
\[ S' = \frac{1.4m u^2}{2 F_b} \]

7. **Relate \( S' \) to \( S \)**:
- From the earlier expression for \( S \):
\[ S' = 1.4 \cdot S \]
- Thus, the new stopping distance \( S' \) is:
\[ S' = \frac{7}{5} S = 1.4 S \]

### Conclusion:
The stopping distance when passengers add 40% to the weight of the car is \( 1.4S \).