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# A train $110$ meters long is running with a speed of $60$ kmph. In what time will it pass a man who is running at $6$ kmph in the direction opposite to that in which the train is going?

Last updated date: 15th Jun 2024
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Hint: In basic type of question related to speed, distance and time we have one moving object but when we have two objects and out of them one is for sure moving and the second one is either static or moving and when they are moving either in the same direction or in the opposite direction they will cross each other. Then there comes the concept of relative speed. Relative speed in the case of the opposite direction of movement is added up but in the case of same direction movement it becomes the difference of the two.

Complete step by step solution: If two object bodies are moving in opposite direction at $u$ kmph and $v$ kmph, where$u > v$,
then their relative speed is $= (u + v)$ kmph
Here in our question we have a train and a man as two object bodies.
Therefore on comparing from the above statement the given information can be written as
$u = 60$ kmph, $v = 6$ kmph.
The length of the train, say $l$ =$110$ m
We know the formula for time $t$ taken by a moving object when speed is $s$ and the distance has to be covered is $d$ is $t = \dfrac{d}{s}$.
Now the time taken by the train to cross the man $t = \dfrac{l}{{u + v}}$
Therefore by substituting the values, we have
$t = \dfrac{{110}}{{60 + 6}} = \dfrac{{110}}{{66}} = \dfrac{5}{3} = 1\dfrac{2}{3}h$
So the required time is in mixed fraction form. The fraction part of mixed fraction is in hour and can be converted into minutes.
Hence time required by a train to pass the man running in the opposite direction is 1 hour and40 minutes.

Note: 1. If two object bodies are moving in the same direction at $u$kmph and $v$ kmph, where $u > v$, then their relative speed is = $(u - v)$ kmph.
2. If both the object bodies have remarkable length then the distance has to be covered becomes $d + l_1 + l_2$ where $d$ is the distance between them, $l_1$ is the length of one object and $l_2$ is length of the second object.