Question

A metro trains starts from rest and in $5 \,s$ achieves $108 \,km/h$. After that it moves with constant velocity and comes to rest after travelling $45 \,m$ with uniform retardation. If total distance travelled is $395 \,m$, find total time of travelling.

• A. 12.2s
• B. 15.3s
• C. 9s
• D. 17.2s

Answer 1

Answer: (D)

17.2s

Answer 2

Given $v = 108 \, kmh^{-1} = 30 \,ms^{- 1}$
From first equation of motion
v = u + at
hence $30=0+a\times 5$
(hence $u=0)$
or $a=6\, ms^{-2}$
So, distance travelled by metro train in $5\, s$
$s_1=\frac{1}{2}at^2=\frac{1}{2}\times (6)\times (5)^2=75\, m$
Distance travelled before coming to rest $=45 \,m$
So, from third equation of motion
$o^2=(30)^2-2a' \times 45$
or a' $=\frac{30 \times 30 }{2 \times 45 }=10\, ms^{-2}$
Time taken in travelling $45 \,m$ is
$t_3=\frac{30}{10}=3\, s$
Now, total distance = 395 ,m
i,e $75 + y + 45 = 395\, m$
or $s'=395 -(75 +45)=275 \,m$
$\therefore$ $t_2=\frac{275}{30}=9.2\,s$
Hence, total time taken in whole journey
$= t_1+t_2+t_3$
$=5+9.2+3$
$=17.2\,s$