Question

A metro trains strats from rest and in five seconds 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, $v=108km/h=30m/s$
From first equation of motion.
$v=u+at$
$\therefore 30=0+a\times 5(\because u=0)$
or $a=6m/s^2$
So, distance travelled by metro train in $5s$
$s_1=\frac{1}{2}a{t}^2$
$=\frac{1}{2}\times \left(6\right)\times {\left(5\right)}^2=75m$
Distance travelled before coming to rest
$=45m$
So, from third equation of motion
$0^2={\left(30\right)}^2-2a^{\prime }\times 45$
or $a^{\prime }=\frac{30\times 30}{2\times 45}=10m/s^2$
Time taken in travelling $45m$ is
$t_3=\frac{30}{10}=3s$
Now, total distance $=395m$
i.e., $75+s^{\prime }+45=395m$
or $sʹ=395-(75+45)=275m$
$\therefore t_2=\frac{275}{30}=9.2s$
Hence, total time taken in whole journey
$=t_1+t_2+t_3$
$=17.2s$