Question

A metro train starts from rest and in $5s$ achieves $108km/hr$. After that it moves with constant velocity and comes to rest after travelling $45m$ with uniform retardation. If the total distance travelled is $395m$, find the total time of travelling.
A. $12.2s$
B. $15.3s$
C. $9s$
D. $17.2s$

Answer 1

Hint Break the journey in three parts and solve for time in each part. Use the three equations of motion in each part according to the information given in the question to find the desired quantities.

Complete step-by-step solution :
We can break the journey in three parts. First part is where the train is accelerating, second part where the train moves with constant velocity, and the third part is where the train is retarding to rest.
For the first part of journey,
${t_1} = 5s$
$u = 0m/s$
$v = 108km/hr = 30m/s$
Using first equation of motion,
$\begin{gathered} v = u + at \\\ a = \dfrac{{v - u}}{t} \\\ a = \dfrac{{30 - 0}}{5} \\\ a = 6m/{s^2} \\\ \end{gathered}$
Using third equation of motion,
${v^2} - {u^2} = 2aS$
$\begin{gathered} S = \dfrac{{{v^2} - {u^2}}}{{2a}} \\\ \\\ \end{gathered}$
$\begin{gathered} S = \dfrac{{{{30}^2} - 0}}{{2 \times 6}} \\\ S = 75m \\\ \end{gathered}$
For second part of journey,
Let the time for which train is moving with constant velocity be ${t_{_2}}$
$S = v{t_2}$
$S = 30{t_2}$
For third part of journey,
$u = 30m/s$
$v = 0m/s$
$S = 45m$
Using third equation of motion,
${v^2} - {u^2} = 2aS$
$a = \dfrac{{{v^2} - {u^2}}}{{2S}}$
$a = \dfrac{{0 - 900}}{{2 \times 45}}$
$a = - 10m/{s^2}$
Using first equation of motion,
$v = u + a{t_3}$
${t_3} = \dfrac{{v - u}}{a}$
${t_3} = \dfrac{{0 - 30}}{{ - 10}}$
${t_3} = 3s$
Now,
It is given in the question that the total journey comprises of $395m$,
Total distance$= 395$
$75 + 30{t_2} + 45 = 395$
${t_2} = 9.16s$
Total time of journey$= {t_1} + {t_2} + {t_3}$
Total time of journey$\begin{gathered} = 5 + 9.16 + 3 \\\ = 17.16s \\\ \approx 17.2s \\\ \end{gathered}$
d)correct.

Note:- We can also do this question from graphical method. For this approach, first we have to plot the velocity time graph of the train and find the area of this velocity-time graph because the area of velocity-time graph gives us the displacement which we can use to find the time for the second part of the journey.