Explanation

(d) Given, $v = 10\ g\ {kmh}^{- 1} = 30\ mh^{- 1}$
From first equation of motion.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = u + at$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 30 = 0 + a \times 5\ \ \ \ \ \ \ \ \ \ \ (\because u = 0)$
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
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0^{2} = (30)^{2} - 2a' \times 45$
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 $= 395\ m$
$ie,\ \ 75 + s^{'} + 45 = 395\ m$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ s^{'} = 395 - (75 + 45) = 275\ m$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t_{2} = \frac{275}{30} = 9.2\ s$
Hence, total time taken in whole journey
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = t_{1} + t_{2} + t_{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5 + 9.2 + 3 = 17.2\ s$