A bus can be stopped by applying a retarding force F when it is moving with a speed v on a level road. The distance covered by it before coming to rest is s. If the load of the bus increases 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

To find the new stopping distance, we can use the work-energy theorem. The work-energy theorem states that the work done by the retarding force on the bus is equal to the change in its kinetic energy.

Initially, let the mass of the bus be $m$ and its initial velocity be $v$.
Since the bus comes to rest, its final velocity is 0.
The work done by the retarding force $F$ over a stopping distance $s$ is given by:
$W=-F\cdot s$

The change in kinetic energy ($\Delta K$) is:
$\Delta K=K_f-K_i=0-\frac{1}{2}m{v}^2$

Equating the work done to the change in kinetic energy:
$-F\cdot s=-\frac{1}{2}m{v}^2$
$F\cdot s=\frac{1}{2}m{v}^2$
From this, we can express the stopping distance $s$ as:
$s=\frac{m{v}^2}{2F}$ (Equation 1)

Now, the load of the bus increases by 50% because of passengers.
The new mass $m^{\text{'}}$ of the bus is:
$m^{\text{'}}=m+0.5m=1.5m$

For the same initial speed $v$ and same retarding force $F$, let the new stopping distance be $s^{\text{'}}$.
Using the same relation for the new scenario:
$s^{\text{'}}=\frac{m^{\text{'}}{v}^2}{2F}$

Substitute $m^{\text{'}}=1.5m$ into the equation:
$s^{\text{'}}=\frac{\left(1.5m\right)v^2}{2F}=1.5\cdot \left(\frac{m{v}^2}{2F}\right)$

From Equation 1, we know that $\frac{m{v}^2}{2F}=s$.
Therefore, we get:
$s^{\text{'}}=1.5s$

Thus, the distance covered by the bus to come to rest shall be 1.5s.