A bus can be stopped by applying a retarding force F when it is moving with speed v on a level road. The distance covered by it before coming to rest is `s`. If the load of the bus increase by 50% because of passengers, for the same speed and same retarding force, the distance covered by the bus to come to rest shall be

Options

A. (a) `1.5s`
B. (b) `2s`
C. (c) `1s`
D. (d) `2.5s`

Explanation

To solve the problem, we need to analyze the relationship between the retarding force, the mass of the bus, and the distance covered before coming to rest.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- Let the mass of the bus be \( m \).
- The initial speed of the bus is \( v \).
- The distance covered before coming to rest is \( s \).
- The retarding force applied is \( F \).

2. **Using Work-Energy Principle**:
- The work done by the retarding force \( F \) is equal to the change in kinetic energy of the bus.
- The initial kinetic energy of the bus is given by:
\[ KE_{initial} = \frac{1}{2} m v^2 \]
- The final kinetic energy when the bus comes to rest is:
\[ KE_{final} = 0 \]
- Therefore, the work done by the retarding force is:
\[ W = KE_{final} - KE_{initial} = 0 - \frac{1}{2} m v^2 = -\frac{1}{2} m v^2 \]

3. **Relating Work Done to Distance**:
- The work done by the retarding force can also be expressed as:
\[ W = F \cdot s \]
- Setting the two expressions for work equal gives:
\[ F \cdot s = \frac{1}{2} m v^2 \]

4. **Considering the Increased Load**:
- If the load of the bus increases by 50%, the new mass \( m' \) becomes:
\[ m' = m + 0.5m = \frac{3m}{2} \]

5. **Calculating the New Distance**:
- Using the same retarding force \( F \) and the same initial speed \( v \), the work done when the mass is increased is:
\[ W' = F \cdot s' = \frac{1}{2} m' v^2 \]
- Substituting \( m' \):
\[ F \cdot s' = \frac{1}{2} \left(\frac{3m}{2}\right) v^2 = \frac{3}{4} m v^2 \]

6. **Setting Up the Ratio**:
- Now we have two equations:
1. \( F \cdot s = \frac{1}{2} m v^2 \)
2. \( F \cdot s' = \frac{3}{4} m v^2 \)
- Dividing the second equation by the first:
\[ \frac{F \cdot s'}{F \cdot s} = \frac{\frac{3}{4} m v^2}{\frac{1}{2} m v^2} \]
- Simplifying gives:
\[ \frac{s'}{s} = \frac{\frac{3}{4}}{\frac{1}{2}} = \frac{3}{2} \]

7. **Final Result**:
- Therefore, the new distance \( s' \) is:
\[ s' = \frac{3}{2} s \]

### Conclusion:
The distance covered by the bus to come to rest after the load increases by 50% is \( \frac{3}{2} s \) or \( 1.5s \).