Question

An unloaded bus can be stopped by applying brakes on a straight road after covering a distance x. Suppose, the passenger adds 50% of its weight as the load and the breaking force remains unchanged, how far will the bus go after the application of brakes? (Velocity of bus in each case is same)
a) x
b) 1.5x
c) 2x
d) 2.5x

Answer 1

Hint: In this case the velocity of the bus remains unchanged. The force with which brakes are applied is also the same. The only factor that needs to be kept in mind is the change in weight(mass) of the bus because of the passenger. After the passenger boards the bus the mass of the bus changes to 100% + 50% = 150%. This means that the final mass of the bus becomes 1.5 times the initial mass of the bus. So, by using newton’s law: $F=ma$ , find the final acceleration. Then, by using the equation of motion: ${{v}^{2}}={{u}^{2}}+2as$, find the displacement.

Formula used:
${{v}^{2}}={{u}^{2}}+2as$, where v is final velocity, u is initial velocity, a is acceleration and s is the distance covered.
$F=ma$, where F is force, m is mass of the object and a is the acceleration.

Complete step by step answer:
As we have:
$\begin{align}
  & v=0 \\
 & u=\text{constant} \\
 & s=x \\
\end{align}$
So, by equation of motion:
${{v}^{2}}={{u}^{2}}+2as$
We have:
$s=\dfrac{{{u}^{2}}}{2a}$
So, we can say that:
$s\propto \dfrac{1}{a}$
i.e. $x=\dfrac{1}{a}......(1)$
Also, we know that:
$F=\text{constant}$
So, we can say that:
\[\begin{align}
  & \Rightarrow {{m}_{i}}{{a}_{i}}={{m}_{f}}{{a}_{f}} \\
 & \Rightarrow ma=1.5m{{a}_{f}} \\
 & \Rightarrow {{a}_{f}}=\dfrac{1}{1.5} \\
\end{align}\]
So, we get:
$
{{s}_{f}}=\dfrac{1}{\dfrac{1}{1.5}a} \\
\implies {{s}_{f}}=1.5\dfrac{1}{a} \\
\therefore {{s}_{f}}=1.5x \\
$

So, the correct answer is “Option B”.

Note:
Equations of motion must be kept in mind while solving the problems based on motion of bodies. There are three general equations of motion. This question requires us to remember the relation between distance and acceleration of a body.
According to the third equation of motion: ${{v}^{2}}={{u}^{2}}+2as$ , distance is inversely proportional to acceleration of body. So, when retardation a becomes $\dfrac{1}{1.5}$ times, then the distance will become inverse of it, i.e. 1.5 times.