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# A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/hr. The other one walks at 5.4 km/ hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train? A.$66\text{ km/hr}$B. $\text{72 km/hr}$C. $78\text{ km/hr}$D. $81\text{ km/hr}$

Last updated date: 19th Sep 2024
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Hint: We use the fact that if two objects are moving in the same direction with speeds say ${{v}_{1}},{{v}_{2}}$ then their relative speed is given by ${{v}_{1}}+{{v}_{2}}$. We assume the speed of the train as $x$ m/sec Since the train to has to cover the distance of the own length to cross the men we have $l={{r}_{1}}{{t}_{1}}={{r}_{2}}{{t}_{2}}$ where ${{r}_{1}},{{r}_{2}}$ are relative speed of the train with respect to the mean d ${{t}_{1}},{{t}_{2}}$ are the time to cross them. We solve for $x$. 

Complete step-by-step solution:
Let us speed the first man and second man as ${{v}_{1}},{{v}_{2}}$ m/sec respectively. We denote the time taken by the train to cross the first man as ${{t}_{1}}$ sec and the second man as ${{t}_{2}}$ sec. We are give in the question
\begin{align} & {{v}_{1}}=4.5\text{ k/hr}=4.5\times \dfrac{5}{18}\text{ m/sec}=\dfrac{.25}{2}\text{ m/sec}=1.25\text{ m/sec} \\ & {{v}_{2}}=5.4\text{ k/hr}=5.4\times \dfrac{5}{18}\text{ m/sec}=\dfrac{.3}{2}\text{ m/sec}=1.5\text{ m/sec} \\ \end{align}
Let us assume the speed of the train is $x$ m/sec. We denote the relative speed of the train with respect to first man as ${{r}_{1}}$ and second man as${{r}_{2}}$. So we have
\begin{align} & {{r}_{1}}=x-{{v}_{1}}=x-1.25\text{ m/sec } \\ & {{r}_{1}}=x-{{v}_{1}}=x-1.5\text{ m/sec} \\ \end{align}
We know that the train has to cross a distance equal to its own length to cross the men. Let us denote the length of the train as $l$. So the distance covered by the train while crossing the first man is
$l={{r}_{1}}{{t}_{1}}$
Similarly the distance covered by the train while crossing the second man is
$l={{r}_{2}}{{t}_{2}}$
So we have
\begin{align} & {{r}_{1}}{{t}_{1}}={{r}_{2}}{{t}_{2}} \\ & \Rightarrow \left( x-1.25 \right)\times 8.4=\left( x-1.5 \right)\times 8.5 \\ & \Rightarrow 8.4x-10.5=8.5x-12.75 \\ & \Rightarrow 0.1x=2.25 \\ & \Rightarrow x=22.5\text{ m/sec} \\ \end{align}
So the speed train in km/hr is
$x=22.5\times \dfrac{18}{5}=\dfrac{45}{2}\times \dfrac{18}{5}=81\text{ km/hr}$
So the correct option is D.

Note: We should remember the conversion between km/hr and m/sec as $1\text{ km/hr}=\dfrac{5}{18}\text{ m/sec}$ and $1\text{ m/sec}=\dfrac{18}{5}\text{km/hr}$. We can solve it without converting but calculation will be difficult. We note that if two objects move in opposite directions their relative speed is given by ${{v}_{1}}+{{v}_{2}}$ . We have assumed the length of the man as negligible here but the train has to cross a platform or bridge of; length $d$ then it has to cross $l+d$ distance to cross the platform or the bridge completely. The distance covered by an object which speed $v$ and time $t$ is given by $vt$.