Question

A metro trains strats from rest and in five seconds 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. $ 12.2\,s $
• B. $ 15.3\,s $
• C. $ 9\, s $
• D. $ 17.2\,s $

Answer 1

Answer: (D)

Given, $v =108\,km/h = 30\,m/s$
From first equation of motion.
$v = u + at$
$∴ 30 = 0 + a × 5 \quad ( \because u = 0)$
or $a = 6 \,m/s^2$
So, distance travelled by metro train in $5\,s$
$s_{1} = \frac{1}{2} at^{2} $
$= \frac{1}{2}\times\left(6\right) \times\left(5\right)^{2} = 75\, m$
Distance travelled before coming to rest
$= 45 \,m$
So, from third equation of motion
$ 0^{2} = \left(30\right)^{2} -2a'\times 45$
or $ a' =\frac{ 30 \times 30}{2\times 45}=10 m/s^{2} $
Time taken in travelling $45\,m$ is
$t_{3} = \frac{30}{10} = 3\, s $
Now, total distance $= 395\,m$
i.e., $75 + s' + 45 = 395\, m$
or $sʹ = 395 - ( 75 + 45) = 275 \,m $
$ ∴ t_2 = \frac{275}{30} = 9 .2\, s$
Hence, total time taken in whole journey
$ = t_1 +t_2+ t_3$
$= 17 . 2\, s$