Question

Find the time taken by a train 150 meters long, running at \[60km/hr\] in crossing a man standing on a platform.

Answer 1

Hint- In this question, the train is moving while the man is at rest at the platform so the relative velocity between the train and the man is the same as the velocity of the train. We need to determine the time taken by the train to pass the man who is standing on the platform for which we have to use the formula $ {\text{Distance = velocity x time}} $.

Complete step by step solution:

Let the time taken by the train to pass by a man who is standing on a platform be $ t{\text{ seconds}}{\text{.}} $

As the train is 150 meters long and is moving at a speed of \[60km/hr\]with respect to the man standing on the platform, the relative speed between the train and the man is \[60km/hr\].

First, we need to first convert 150 meters to kilometers to get all the measuring units in one dimension only. To convert meters to kilometers, divide the numeric of meters by 1000 we get,
$
  150{\text{ }}meters = \dfrac{{150}}{{1000}}{\text{ kilometers}} \\
  {\text{ = 0}}{\text{.15 kilometers}} \\
 $

According to the question, the total distance traveled by the train is 0.15 kilometers to pass the man with the speed of \[60km/hr\]is $ t{\text{ seconds}}{\text{.}} $ Mathematically,

$
  0.15 = 60 \times t \\
  t = \dfrac{{0.15}}{{60}}{\text{ hours}} \\
  {\text{ = 2}}{\text{.5}} \times {10^{ - 3}}{\text{ hours}} \\
  {\text{ = 2}}{\text{.5}} \times {10^{ - 3}} \times 60{\text{ minutes}} \\
  {\text{ = 2}}{\text{.5}} \times {10^{ - 3}} \times 60 \times 60{\text{ seconds}} \\
  {\text{ = 9 seconds}} \\
 $

Hence, the time taken by the train 150 meters long, running at \[60km/hr\] in crossing a man standing on a platform is 9 seconds.

Note: It is to be noted here that, before starting to solve the question, all the given quantities or, numerical values should be in the same measuring dimensions. Here, we have converted meters to kilometers while kilometer per hour can also be converted to meter per second.