Two trains start at the same time from two stations and proceed towards each other with speeds 40 km per hour and 45 km per hour respectively. When they meet, it is found that one train has travelled 40 km more than the other. Find the distance between the two stations \[\]
A.360 km \[\]
B.680 km \[\]
C.720 km\[\]
D.550 km \[\]

Hint: We assume that the slower train ${{T}_{1}}$ with speed ${{v}_{1}}=40$km/hr has travelled $x$ km when it meets the faster train${{T}_{2}}$. Hence the distance travelled by train ${{T}_{2}}$ when it meets ${{T}_{1}}$is $\left( x+40 \right)$km. We find the tame take by train ${{T}_{1}}$ and ${{T}_{2}}$as ${{t}_{1}},{{t}_{2}}$ to meet in between the station and equate ${{t}_{1}}={{t}_{2}}$ to find $x,x+40$ and the distance between trains as the sum of distances travelled by trains $x+x+40$.\[\]

Complete step-by-step answer:
We know the distance covered $d$ by an object is the length of shortest possible path from the initial position to the latter position. If we measure the time from point of start as $t$ then the speed $v$ is the distance covered per unit time. It is given by the formula
\[v=\dfrac{s}{t}\]

We are given the question that two trains start at the same time from two stations and proceed towards each other with speeds 40 km per hour and 45 km per hour respectively. Let us denote the first train as ${{T}_{1}}$ and speed of the first train as ${{v}_{1}}$. Similarly we denote the second train as ${{T}_{2}}$ and speed of train ${{T}_{2}}$ as ${{v}_{2}}$. We have,
\[{{v}_{1}}=40\text{km/hr},{{v}_{2}}=45\text{km/hr}\]
We are further given the question that if the two trains meet it is found that one train has travelled 40 km more than the other. Let us assume that the slower train ${{T}_{1}}$ has travelled $x$ km when it meets the faster train${{T}_{2}}$. Hence the distance travelled by train ${{T}_{2}}$ is $\left( x+40 \right)$km. The time taken ${{t}_{1}}$ by slower train ${{T}_{1}}$ to meet faster train ${{T}_{2}}$ is
\[{{t}_{1}}=\dfrac{x}{{{v}_{1}}}=\dfrac{x}{40}\text{hr}\]
The time taken ${{t}_{1}}$ by faster train ${{T}_{2}}$ to meet faster train ${{T}_{1}}$ is
\[{{t}_{2}}=\dfrac{x+40}{{{v}_{2}}}=\dfrac{x+40}{45}\text{hr}\]
We are given both the trains ${{T}_{1}},{{T}_{2}}$ start at the same time which means they take equal time to meet each other. So we have
\[\begin{align}
  & {{t}_{1}}={{t}_{2}} \\
 & \Rightarrow \dfrac{x}{40}=\dfrac{x+40}{45} \\
 & \Rightarrow 45x=40x+1600 \\
 & \Rightarrow 5x=1600 \\
 & \Rightarrow x=320 \\
\end{align}\]
So the distance travelled by ${{T}_{1}}$ is $x=320$km and the distance travelled by ${{T}_{2}}$is $x+40=320+40=360$km. The distance between the stations as the sum of distances travelled by trains which is?
\[x+x+40=320+320+40=680\text{km}\]
So the correct option is B. \[\]

So, the correct answer is “Option B”.

Note: We note that the key here is time remaining the same for both the trains. The relative speed of trains moving opposite direction with speed ${{v}_{1}},{{v}_{2}}$ is given by ${{v}_{1}}+{{v}_{2}}$. If a train crosses a platform of length ${{l}_{p}}$ then the train has to also cover its own length ${{l}_{t}}$ and the total distance travelled will be ${{l}_{p}}+{{l}_{t}}$.