Question

The speed of the train is reduced from 36km per hour to 9km per hour whilst it travels a distance of 150metres, if the retardation be uniform, find how much further it will travel before coming to rest.
A. 10m
B. 12m
C. 14m
D. 16m

Answer 1

In the question we are given two parts of motion: one where the speed is reduced from 36km per hour to 9km per hour while covering 150m distance and other where the train comes to rest. From the first part we could get the acceleration of the train. Using the same you could find the distance covered before coming to rest.
Formula used:
Equation of motion,
${{v}^{2}}-{{u}^{2}}=2as$

Complete step-by-step solution:
In the question, we are given the speed of the train to reduced from 36km per hour to 9km per hour when travelling a distance of 150meters. We are supposed to find the distance that the train covers before coming to rest assuming the retardation to be uniform.
So in the first part we have,
$u=36\times \dfrac{5}{18}=10m{{s}^{-1}}$
$v=9\times \dfrac{5}{18}=2.5m{{s}^{-1}}$
S=150m
Now we have the equation of motion that is given by,
${{v}^{2}}-{{u}^{2}}=2as$
Substituting the values we get,
$a=\dfrac{100-6.25}{150\times 2}=0.3125m{{s}^{-2}}$
Now for the second part we have,
$u=2.5m{{s}^{-1}}$
V=0
Now the equation motion in the second part of the question will become as following if we substitute the acceleration found above,
$s=\dfrac{6.25}{2\times 0.3125}=10m$
Therefore, we found that the train will move a distance of 10m before coming to rest. Hence, option A is found to be the correct answer.

Note: In the question, we are given the velocity of the train km per hour which is not the SI unit of velocity. Since all the other quantities are given in SI units, we will have to convert all of them to SI units. So we have multiplied the given value by $\dfrac{5}{18}$ so as to convert km per hour into meters per second.