A car without passengers moving with certain velocity on a level ground can be stopped in a distance of 10 m. If the passengers add 25% of its weight, its stopping distance for the same braking force and velocity is (ignore (friction)

Options

A. 15m
B. 10m
C. 7.5m
D. 12.5m

Answer

To solve the problem, we will analyze the stopping distance of a car with and without passengers, using the principles of physics related to work, energy, and motion.

### Step-by-Step Solution:

1. **Define the Initial Scenario (Case 1)**:
- Let the mass of the car without passengers be \( m \).
- The car can stop from an initial velocity \( u \) to a final velocity \( v = 0 \) in a distance \( s_1 = 10 \, \text{m} \).
- The braking force \( F \) is constant.

2. **Use the Equation of Motion**:
- We apply the third equation of motion:
\[ v^2 = u^2 + 2as \]
- Substituting \( v = 0 \) and \( s = s_1 = 10 \):
\[ 0 = u^2 + 2a(10) \]
- Rearranging gives:
\[ u^2 = -20a \]

3. **Determine the Acceleration**:
- The braking force \( F \) provides the acceleration \( a \):
\[ a = \frac{F}{m} \]
- Substituting this into the equation:
\[ u^2 = -20 \left(\frac{F}{m}\right) \]

4. **Define the New Scenario (Case 2)**:
- When passengers are added, the new mass \( m' \) becomes:
\[ m' = m + 0.25m = 1.25m = \frac{5m}{4} \]
- The braking force remains \( F \).

5. **Calculate the New Acceleration**:
- The new acceleration \( a' \) is:
\[ a' = \frac{F}{m'} = \frac{F}{\frac{5m}{4}} = \frac{4F}{5m} \]

6. **Apply the Equation of Motion for Case 2**:
- Again using the third equation of motion:
\[ 0 = u^2 + 2a's \]
- Substituting \( a' \):
\[ 0 = u^2 + 2\left(\frac{4F}{5m}\right)s \]
- Rearranging gives:
\[ u^2 = -\frac{8Fs}{5m} \]

7. **Equate the Two Expressions for \( u^2 \)**:
- From Case 1:
\[ u^2 = -20\left(\frac{F}{m}\right) \]
- From Case 2:
\[ u^2 = -\frac{8Fs}{5m} \]
- Setting them equal:
\[ -20\left(\frac{F}{m}\right) = -\frac{8Fs}{5m} \]

8. **Solve for the Stopping Distance \( s \)**:
- Cancel \( F \) and \( m \):
\[ 20 = \frac{8s}{5} \]
- Rearranging gives:
\[ s = \frac{20 \cdot 5}{8} = \frac{100}{8} = 12.5 \, \text{m} \]

### Final Answer:
The stopping distance for the car with passengers is **12.5 meters**.