Question

How time is required to bring a train to rest if it takes 5sec to reduce its speed from 90kmph to 36kmph
a) 5s
b) 8.33s
c) 10.2s
d) 12.43s

Answer 1

Hint: It is given to us that the train is decelerating i.e. its speed gradually reduces with time. Let us say the train is decelerating uniformly. We will first calculate the rate of deceleration from Newton’s first kinematic equation from the given data. Further taking 36kmph as our initial speed we will calculate the time taken for the train to come to rest from that position.
Formula used:
$V=U+at$

Complete step-by-step solution:
Let us say that a body moves with an acceleration ’a’. if the initial velocity of the body is ‘U’ than its velocity at time ‘t’ is given by,
$V=U+at$
If the body is decelerating then the declaration is equal to ‘-a’.
In the above question it is given to us that the train takes 5sec to reduce its speed from 90kmph to 36kmph. Hence the declaration from the above Newton’s first kinematic equation we get as,
$\begin{align}
  & V=U+at \\
 & \Rightarrow 36=90+a5 \\
 & \Rightarrow 5a=36-90 \\
 & \Rightarrow a=-\dfrac{54}{5}=-10.8m{{s}^{-2}} \\
\end{align}$
Let us say from this instant of time the train comes to rest. Therefore we can say that the final velocity of the train is zero. Hence from the above Newton’s first kinematic equation we get the time taken for the train to come to rest from speed of 30kmph as,
$\begin{align}
  & V=U+at \\
 & \Rightarrow 0=36+(-10.8)t \\
 & \Rightarrow -36=-10.8t \\
 & \Rightarrow t=\dfrac{36}{10.8}=3.33\sec \\
\end{align}$
Hence the total time taken for the train to come to rest is 5 sec plus 3.33 sec i.e. equal to 8.33sec. Hence the correct answer of the above question is option b.

Note: We have assumed that the train uniformly reduces its speed with time. Therefore the above equations used to hold its validity. The minus sign next to acceleration indicated that there is a force that acts in the opposite direction to the motion of the train, which eventually brings it to rest.