Q115 / 160 · JIPMER 2011 · Physics

A metro train starts from rest and in 5 s achieves $108 \mathrm{~km} / \mathrm{h}$. 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
  • Correct answer
    D
    17.2 s

Solution

Given, $v=108 \mathrm{~km} / \mathrm{h}=30 \mathrm{~m} / \mathrm{s}$ For first equation of motion $v=u+a t$ $\therefore \quad 30=0+a \times 5 \quad(\because u=0)$ or $\quad a=6 \mathrm{~m} / \mathrm{s}^2$ So, distance travelled by metro train in 5 s $s_1=\frac{1}{2} a t^2=\frac{1}{2} \times(6) \times(5)^2=75 \mathrm{~m}$ Distance travelled before coming to rest $=45 \mathrm{~m}$ So, from third equation of motion $\begin{aligned} & 0^2=(30)^2-2 a^{\prime} \times 45 \\ & \text { or } \quad a^{\prime}=\frac{30 \times 30}{2 \times 45}=10 \mathrm{~m} / \mathrm{s}^2 \\ & \end{aligned}$ Time taken in travelling 45 m is $t_3=\frac{30}{10}=3 \mathrm{~s}$ Now, total distance $=395 \mathrm{~m}$ ie, $\quad 75+s^{\prime}+45=395 \mathrm{~m}$ or $s^{\prime}=395-(75+45)=275 \mathrm{~m}$ $\therefore \quad t_2=\frac{275}{30}=9.2 \mathrm{~s}$ Hence, total time taken in whole journey $\begin{aligned} & =t_1+t_2+t_3 \\ & =5+9.2+3 \\ & =17.2 \mathrm{~s}\end{aligned}$

Hence, the correct answer is option: 'D'.

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