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Question: A train \(110\,m\) in length travels at \(60\) per hour. In what time it will pass a man who is walk...

A train 110 m110\,m in length travels at 6060 per hour. In what time it will pass a man who is walking at 6 km6\,km an hour in the opposite direction.
A. 5sec⁡5\sec
B. 6sec⁡6\sec
C. 8sec⁡8\sec
D. 713sec⁡7\dfrac{1}{3}\sec

Explanation

Solution

In order to solve this question we need to understand the relative motion. To define relative motion, we first learn a frame of reference. So a frame of reference is defined like a window or perspective in which the event is observed. So a relative motion between two bodies is defined as the motion which is observed with respect to one another, it has two categories, one in which newton’s law valid or when the speed of two bodies are very less in comparison to speed of light, and the second category in which Lorentz transformation valid or when the speed of two bodies are comparable to light speed.

Complete step by step answer:
According to the question, the length of the train is given as, L=110 mL = 110\,m. Let the train moving in direction of positive x-axis, so its velocity with respect to ground is given as,
V⃗TG=60 kmhr−1i^{\vec V_{TG}} = 60\,kmh{r^{ - 1}}\hat i
⇒V⃗TG=60×518msec⁡−1i^\Rightarrow {\vec V_{TG}} = 60 \times \dfrac{5}{{18}}m{\sec ^{ - 1}}\hat i
⇒V⃗TG=16.67msec⁡−1i^\Rightarrow {\vec V_{TG}} = 16.67m{\sec ^{ - 1}}\hat i
Also, let the man moving in opposite direction, so its velocity with respect to ground is given as,
V⃗MG=−6kmhr−1i^{\vec V_{MG}} = - 6kmh{r^{ - 1}}\hat i
⇒V⃗MG=−6×518msec⁡−1i^\Rightarrow {\vec V_{MG}} = - 6 \times \dfrac{5}{{18}}m{\sec ^{ - 1}}\hat i
⇒V⃗MG=−1.67 msec⁡−1i^\Rightarrow {\vec V_{MG}} = - 1.67\,m{\sec ^{ - 1}}\hat i

So velocity of train with respect to man is given as,
V⃗TM=V⃗TG−V⃗MG{\vec V_{TM}} = {\vec V_{TG}} - {\vec V_{MG}}
Putting values we get,
V⃗TM=[16.67i^−(−1.67i^)]msec⁡−1{\vec V_{TM}} = [16.67\hat i - ( - 1.67\hat i)]m{\sec ^{ - 1}}
⇒V⃗TM=(16.67+1.67)i^msec⁡−1\Rightarrow {\vec V_{TM}} = (16.67 + 1.67)\hat im{\sec ^{ - 1}}
⇒V⃗TM=18.34i^msec⁡−1\Rightarrow {\vec V_{TM}} = 18.34\hat im{\sec ^{ - 1}}
Let the time taken by the train to cross a man be, tt
Since we know, ∣v⃗∣=dt\left| {\vec v} \right| = \dfrac{d}{t}
So time period by train to cross men is,
t=L∣V⃗TM∣t = \dfrac{L}{{\left| {{{\vec V}_{TM}}} \right|}}
Putting value we get,
t=110 m18.34 m sec⁡−1t = \dfrac{{110\,m}}{{18.34\,m\,{{\sec }^{ - 1}}}}
⇒t=5.99sec⁡\Rightarrow t = 5.99\sec
∴t∼6sec⁡\therefore t \sim 6\sec

So correct option is B.

Note: It should be remembered that here we have neglected the relativistic effect because the speed of both the train and the man is very much less in comparison to speed of train. By relative effect, we physically mean that if one of two bodies is considered to be at rest then what will be its speed in the frame of reference of the other body.