Answer
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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 $vt$.
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 $vt$.
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