Question

A train is running at a speed of $45\dfrac{{km}}{{hr}}$ and a man is walking at a speed of $5\dfrac{{km}}{{hr}}$ in the opposite direction. If the train crosses the man in $18$ seconds then what is its length?
A.$200m$
B.$220m$
C.$180m$
D.$250m$

Answer 1

Hint: Approaching to the problem can be done in such a way that we start with finding the relative speed of the train by adding the speeds of the train and the man as they are moving in the opposite direction then, converting the relative speed in $\dfrac{{km}}{{hr}}$ to $\dfrac{m}{{\sec }}$ using the conversion units.

Complete step by step solution:
We are given that a train is running at a $45\dfrac{{km}}{{hr}}$ and a man is walking at a speed of $5\dfrac{{km}}{{hr}}$ in the opposite direction.
Since, the train and the man are moving in opposite directions, the relative speed of the train is equal to the sum of the speed of the train and the man itself.
So, adding the speed of the train and the man we get:
Relative speed of the train
$ = \left( {45 + 5} \right)\dfrac{{km}}{{hr}} = 50\dfrac{{km}}{{hr}}$.
The value $50\dfrac{{km}}{{hr}}$ can be written as $\dfrac{{50km}}{{1hr}}$.
Since, we know that $1km = 1000m$ that gives $50km = 50 \times 1000m$.
And, we know that $1hr = 60\min $ and $1\min = 60\sec $ that gives $1hr = 60 \times 60\sec $.
Substituting these values in $\dfrac{{50km}}{{1hr}}$ , we get the relative speed as:
$\dfrac{{50km}}{{1hr}} = \dfrac{{50 \times 1000m}}{{60 \times 60\sec }}$.
Solving the right side of the equation, we get:
$\dfrac{{50 \times 1000m}}{{60 \times 60\sec }} = \dfrac{{50 \times 5}}{{18}}\dfrac{m}{{\sec }}$
$ \Rightarrow \dfrac{{50 \times 1000m}}{{60 \times 60\sec }} = \dfrac{{250}}{{18}}\dfrac{m}{{\sec }}$
Therefore, the relative speed of the train becomes:
 $\dfrac{{250}}{{18}}\dfrac{m}{{\sec }}$
Since, we know that the length of the train is equal to the product of relative speed of the train and the time taken by the train to cross the man $\left( {18\sec } \right)$ , which in equation written as:
Length of the train $ = \left( {\dfrac{{250}}{{18}} \times 18} \right)\dfrac{m}{{\sec }} \times \sec $
Solving the brackets, we get:
Length of the train$ = 250m$, which matches the option D:
Therefore, the length of the train is $250m$ and Option D is correct.
So, the correct answer is “Option D”.

Note: If the man would be walking in the same direction as of the train then, we would have subtracted his speed from the speed of the train, as the actual speed for the train would be less than what the speed felt by the man in the opposite direction.