Question

A metro train starts from rest and in $5s$ achieves $108km{h}^{-1}$ After that, it moves with constant velocity and comes to rest after travelling $45m$ with uniform retardation. If the total distance travelled is $395m$, find the total time of travelling.

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

Answer 1

The correct option is D

$17.2s$

The explanation for the correct option:

Step 1:Finding the acceleration
Given

$\begin{array}{rcl}v& =& 108km/h\\ & =& 30m/s\end{array}$

initially at rest $u=0$

From first equation of motion

$\begin{array}{rcl}\mathbf{}\mathit{v}\mathbf{}& \mathbf{=}& \mathbf{}\mathit{u}\mathbf{}\mathbf{+}\mathbf{}\mathit{a}\mathit{t}\\ 30& =& 0+\left(a\times 5\right)(\because \sin ceu=0)\\ a& =& 6m/s^2\end{array}$

Step 2: Finding the distance travelled by metro train in $5s$:

So, the distance travelled by metro train in $5s$

$\begin{array}{rcl}{\mathit{s}}_{\mathbf{1}}& \mathbf{=}& \mathit{u}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathit{b}\mathit{u}\mathit{t}\mathbf{}\mathit{u}\mathbf{}\mathbf{=}\mathbf{0}\\ \mathbf{}\mathbf{}\mathbf{}{\mathit{s}}_{\mathbf{1}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}\mathbf{}\mathbf{}\\ & =& \frac{1}{2}\times \left(6\right)\times {\left(5\right)}^2=75m\end{array}$

Step 3: Finding the new acceleration:

Distance travelled before coming to rest $=45m$

So, from third equation of motion

$\begin{array}{rcl}{u}^2& =& v^2-2as\\ 0& =& {\left(30\right)}^2-2\left(a\text{'}\times 45\right)\\ a\text{'}& =& \frac{30\times 30}{2\times 45}\\ & =& 10m{s}^{-2}\end{array}$

Step 4: Finding the total distance and total time taken:

Time taken in travelling $45m$ is

$\begin{array}{rcl}{t}_3& =& \frac{30}{10}\\ & =& 3s\end{array}$

Now, total distance $=395m$

so,

$\begin{array}{rcl}75+y+45& =& 395mor\\ s\text{'}& =& 395-(75+45)\\ & =& 275m\\ \therefore t_2& =& \frac{275}{30}\\ & =& 9.2s\end{array}$

Hence, total time taken in whole journey

$$\begin{gathered} =t_1+t_2+t_3 \\ =5+9.2+3 \\ =17.2s \end{gathered}$$

Hence, the correct answer is option (D).