Question

A train, 240 m long crosses a man walking along the line in the opposite direction at the rate of 3 kmph in 10 seconds. The speed of the train is
(A) 63 kmph
(B) 75 kmph
(C) 83.5 kmph
(D) 86.4 kmph

Answer 1

Hint: Assume that the speed of the train is x kmph. The speed of the man is 3 kmph. We know that when two objects move towards each other then their relative speed is the summation of their speeds. So, the relative speed between the man and the train is \[\left( x+3 \right)\] kmph. Now, convert the unit of the speed of the train in meters per second using the relation, \[1km=1000m\] and \[1hr=3600\sec \] . The relative distance is 240 meters and the time is 10 seconds. Now, use the formula, \[\text{Relative speed=}\dfrac{\text{Relative distance}}{\text{Time}}\] and get the value of x.

Complete step-by-step answer:
According to the question, it is given that a train, 240 m long crosses a man walking along the line in the opposite direction at the rate of 3 kmph in 10 seconds.
The speed of the man = 3 kmph ………………………….(1)
The time taken by the train to cross the man = 10 seconds ……………………..(2)
The length of the train = 240 meters ……………………………….(3)
Let us assume that the speed of the train is x kmph.
The speed of the train = x kmph …………………………..(4)
We know that when two objects move towards each other then their relative speed is the summation of their speeds.
From equation (1) and equation (4), we have the speed of the train and the man.
Here, the train and the man is moving towards each other. So, the relative speed between them is the summation of their speeds.
The relative speed between the man and the train = \[\left( x+3 \right)\] kmph …………………………..(5)
To cross the man the last point of the train must cross the man. It means that the train has to cover a distance of 240 m.
The relative distance between the man and the train = 240 meters ………………………………(6)
We know that, \[1km=1000m\] ………………………(7)
We also know that, \[1hr=3600\sec \] ………………………………(8)
Now, from equation (5), equation (7), and equation (8), we get
The relative speed between the man and the train = \[\left( x+3 \right)\] kmph = \[\dfrac{\left( x+3 \right)km}{hr}\] = \[\dfrac{\left( x+3 \right)1000m}{3600\sec }=\dfrac{5\left( x+3 \right)}{18}m/s\] ………………………………(9)
We know the formula, \[\text{Relative speed=}\dfrac{\text{Relative distance}}{\text{Time}}\] ……………………………..(10)
Now, from equation (2), equation (6), and equation (10), we get
\[\begin{align}
  & \Rightarrow \dfrac{5\left( x+3 \right)}{18}\text{=}\dfrac{240}{10} \\
 & \Rightarrow 5\left( x+3 \right)=24\times 18 \\
 & \Rightarrow 5x+15=432 \\
 & \Rightarrow 5x=417 \\
 & \Rightarrow x=\dfrac{417}{5} \\
\end{align}\]
The speed of the train =\[\dfrac{417}{5}km/hr=83.4km/hr\] .
Hence, the correct option is (C).

Note:In this question, one might use the formula, \[\text{Relative speed=}\dfrac{\text{Relative distance}}{\text{Time}}\] and then transform it as \[\left( x+3 \right)\text{=}\dfrac{240}{10}\] . This is wrong because the unit of the distance is in meters, time is in seconds, and the speed is in km/hr. The units in the LHS and RHS of the equation \[\left( x+3 \right)\text{=}\dfrac{240}{10}\] , are not the same. So, we have to convert the unit of the speed into meters per second using the relation \[1km=1000m\] and \[1hr=3600\sec \] .