An unloaded bus can be stopped by applying brakies on straight road after covering a distance x. Suppose, the passenger add `50%` of its weight as the load and the braking force remains unchanged, how far will the bus go after the application of the brakes ?(Velocity of bus in both case is same) (Consider negligible friction):-

Options

A. Zero
B. 1.5x
C. 2x
D. 2.5x

Answer

To solve the problem, we need to analyze the situation using the principles of physics, specifically the equations of motion and Newton's second law.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the mass of the unloaded bus be \( m \).
- The bus can stop after covering a distance \( x \) when it is unloaded.
- The initial velocity of the bus is \( u \) and the final velocity when it stops is \( 0 \).

2. **Apply the Equation of Motion**:
- We can use the equation of motion:
\[ v^2 - u^2 = 2as \]
- Here, \( v = 0 \) (final velocity), \( u \) is the initial velocity, \( a \) is the acceleration (which will be negative since it is deceleration), and \( s \) is the distance \( x \).

3. **Express Acceleration in Terms of Force**:
- According to Newton's second law, the acceleration \( a \) can be expressed as:
\[ a = \frac{F}{m} \]
- Where \( F \) is the braking force applied, and \( m \) is the mass of the bus.

4. **Substituting the Values**:
- Substitute \( v = 0 \) and \( a = \frac{F}{m} \) into the equation of motion:
\[ 0 - u^2 = 2 \left(\frac{F}{m}\right)x \]
- Rearranging gives:
\[ -mu^2 = 2Fx \]

5. **Consider the Loaded Bus**:
- When passengers add a load of \( 50\% \) of the bus's weight, the new mass \( m' \) becomes:
\[ m' = m + 0.5m = 1.5m \]
- The braking force \( F \) remains unchanged.

6. **Calculate the New Stopping Distance**:
- Let \( x' \) be the distance the loaded bus will travel before stopping. Using the same logic:
\[ -m'u^2 = 2Fx' \]
- Substitute \( m' = 1.5m \):
\[ -1.5mu^2 = 2Fx' \]

7. **Relate the Two Distances**:
- We can relate the two equations:
\[ -mu^2 = 2Fx \quad \text{(from the unloaded bus)} \]
- Therefore, substituting for \( F \) from the first equation into the second:
\[ -1.5mu^2 = 2\left(-\frac{mu^2}{2x}\right)x' \]
- This simplifies to:
\[ 1.5x = x' \]

8. **Final Result**:
- Thus, the distance \( x' \) that the loaded bus will travel before stopping is:
\[ x' = 1.5x \]

### Conclusion:
The bus will cover a distance of \( 1.5x \) after the application of the brakes when loaded with passengers.