
A train running at the rate of $40\,{\rm{km/h}}$ passes a man riding parallel to the railway line in the same direction at $25\,{\rm{km/h}}$ in $48\,{\rm{seconds}}$. Find the length of the train in metres.
A) ${\rm{100}}\,{\rm{m}}$
B) ${\rm{200}}\,{\rm{m}}$
C) ${\rm{150}}\,{\rm{m}}$
D) ${\rm{250}}\,{\rm{m}}$
Answer
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Hint: In the solution, first we have to calculate the relative speed of the person and the train. Since the train and the person are moving in the opposite direction to each other, we have to add the given speed of the train with the speed of the person. This will give the relative speed. After that we have to apply the formula of the speed \[s = \dfrac{d}{t}\], where $d$ is the distance, $t$ is the time and $s$ is the speed. By substituting the values, we have to calculate the length of the train.
Complete step-by-step answer:
Given,
Speed of the train $ = $$40\,{\rm{km/h}}$
Speed of the man riding parallel to the train $ = $$25\,{\rm{km/h}}$
$\therefore $Speed of the train relative to the man $ = $$40 - 25 = 15\,{\rm{km/h }}$
For the man it appears that the train is going at $15\,{\rm{km/h}}$ with respect to him.
To convert ${\rm{km/h}}$ to ${\rm{m/s}}$we need to multiply the number by $\dfrac{5}{{18}}\,{\rm{m/s}}$.
Converting $15\,{\rm{km/h}}$ into ${\rm{m/s}}$, we get:
$15\,{\rm{km/h = 15}} \times \dfrac{5}{{18}}{\rm{m/s = }}\dfrac{{25}}{6}{\rm{m/s}}$
Time of the train taken to cross the man$ = $$48\,{\rm{seconds}}$
We have to find the length of the train in metres.
So, let the length of the train $ = x$
According to the question,
So, length of the train $ = $(speed with respect to man) $ \times $(time taken to cross the man)
$ = \dfrac{{25}}{6}{\rm{m/s}} \times {\rm{48 seconds = 25}} \times {\rm{8 m = 200 m}}$
Thus, length of the train in metres $ = 200\,{\rm{m}}$
Hence, option B is correct.
Note: Relative speed is a speed where one object is moving with respect to the other. If two objects are moving in the same direction, then their relative speed will be the difference of the speed of the objects. Similarly, when they move in the opposite direction, addition of their speed determines the relative speed. Here we have to determine the total length of the train. Since the speed of the train and the person is given, we can calculate their relative speed. Once we get the relative speed of the person and the train, substituting the values in the formula of speed, we can calculate the length of the train easily.
Complete step-by-step answer:
Given,
Speed of the train $ = $$40\,{\rm{km/h}}$
Speed of the man riding parallel to the train $ = $$25\,{\rm{km/h}}$
$\therefore $Speed of the train relative to the man $ = $$40 - 25 = 15\,{\rm{km/h }}$
For the man it appears that the train is going at $15\,{\rm{km/h}}$ with respect to him.
To convert ${\rm{km/h}}$ to ${\rm{m/s}}$we need to multiply the number by $\dfrac{5}{{18}}\,{\rm{m/s}}$.
Converting $15\,{\rm{km/h}}$ into ${\rm{m/s}}$, we get:
$15\,{\rm{km/h = 15}} \times \dfrac{5}{{18}}{\rm{m/s = }}\dfrac{{25}}{6}{\rm{m/s}}$
Time of the train taken to cross the man$ = $$48\,{\rm{seconds}}$
We have to find the length of the train in metres.
So, let the length of the train $ = x$
According to the question,
So, length of the train $ = $(speed with respect to man) $ \times $(time taken to cross the man)
$ = \dfrac{{25}}{6}{\rm{m/s}} \times {\rm{48 seconds = 25}} \times {\rm{8 m = 200 m}}$
Thus, length of the train in metres $ = 200\,{\rm{m}}$
Hence, option B is correct.
Note: Relative speed is a speed where one object is moving with respect to the other. If two objects are moving in the same direction, then their relative speed will be the difference of the speed of the objects. Similarly, when they move in the opposite direction, addition of their speed determines the relative speed. Here we have to determine the total length of the train. Since the speed of the train and the person is given, we can calculate their relative speed. Once we get the relative speed of the person and the train, substituting the values in the formula of speed, we can calculate the length of the train easily.
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