Question

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

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

Answer 1

Answer: (D)

Given $v=108km{h}^{-1}=30m{s}^{-1}$
From first equation of motion
v = u + at
hence $30=0+a\times 5$
(hence $u=0)$
or $a=6m{s}^{-2}$
So, distance travelled by metro train in $5s$
$s_1=\frac{1}{2}a{t}^2=\frac{1}{2}\times (6)\times (5{)}^2=75m$
Distance travelled before coming to rest $=45m$
So, from third equation of motion
$o^2=(30{)}^2-2a^{\prime }\times 45$
or a' $=\frac{30\times 30}{2\times 45}=10m{s}^{-2}$
Time taken in travelling $45m$ is
$t_3=\frac{30}{10}=3s$
Now, total distance = 395 {\,}m
i,e $75+y+45=395m$
or $s^{\prime }=395-(75+45)=275m$
$\therefore$ $t_2=\frac{275}{30}=9.2s$
Hence, total time taken in whole journey
$=t_1+t_2+t_3$
$=5+9.2+3$
$=17.2s$