An unloaded bus can be stopped by applying brakes on straight road after covering distance x. Suppose. The passengers add `40%` of its weight as the load the braking force remains same. How far will the bus go after the application of the breakes? (Velocity of bus is same in both the cases)

Options

A. x
B. `1.4x`
C. `2x`
D. `2.4x`

Answer

To solve the problem, we will use the work-energy theorem, which states that the work done by all forces acting on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Let the mass of the unloaded bus be \( m \).
- The distance covered by the unloaded bus before stopping is \( x \).
- The initial velocity of the bus is \( v \).

2. **Calculate the Kinetic Energy of the Unloaded Bus:**
- The kinetic energy (KE) of the bus when it is unloaded is given by:
\[ KE = \frac{1}{2} mv^2 \]

3. **Calculate the Work Done by the Braking Force:**
- The work done by the braking force (F) to stop the bus is equal to the kinetic energy:
\[ F \cdot x = \frac{1}{2} mv^2 \]

4. **Determine the New Mass with Passengers:**
- When passengers add 40% of the bus's weight as a load, the new mass \( m' \) becomes:
\[ m' = m + 0.4m = 1.4m \]

5. **Set Up the Equation for the Loaded Bus:**
- The work done to stop the loaded bus (with mass \( m' \)) over a new distance \( x' \) is:
\[ F \cdot x' = \frac{1}{2} m' v^2 \]
- Substituting \( m' \):
\[ F \cdot x' = \frac{1}{2} (1.4m) v^2 \]

6. **Relate the Two Scenarios:**
- From the unloaded scenario, we have:
\[ F \cdot x = \frac{1}{2} mv^2 \]
- Now, substituting this into the equation for the loaded bus:
\[ F \cdot x' = 1.4 \left( F \cdot x \right) \]

7. **Solve for the New Distance \( x' \):**
- Rearranging gives:
\[ x' = 1.4x \]

### Final Answer:
The distance the bus will go after the application of the brakes with the added load is:
\[ x' = 1.4x \]