Calculating Train Velocity When Crossing a Stationary Man
In the field of transportation and physics, understanding the relative speed and velocity of moving objects is crucial. This article delves into the specific scenario where a man standing still at a station crosses a moving train, providing the velocity of the train. We will use various methods to calculate this velocity and discuss the relevant concepts in each approach.Method 1: Using Basic Kinematic Principles
Firstly, we are given a straightforward example where the length of the train is 100 meters and a man crosses this train in 7.2 seconds while the train is traveling in the opposite direction at 5 km/hr. The key to solving this problem lies in understanding the concept of relative speed and time. Let the speed of the train be ( x ) km/hr. The relative speed in terms of meters per second is required to calculate the time taken to cross the train. We convert the speed of the man from km/hr to m/s using the conversion factor ( frac{5}{18} ). Thus, the relative speed is ( x times frac{5}{18} frac{5}{18} frac{5x 5}{18} ) m/s. The total distance to be covered is 100 meters. Using the formula ( text{Time} frac{text{Distance}}{text{Speed}} ), we can set up the equation: [ frac{100}{frac{5x 5}{18}} 7.2 [10pt][ frac{100 times 18}{5x 5} 7.2 [10pt][ frac{1800}{5x 5} 7.2 [10pt][ 1800 7.2 times (5x 5) [10pt][ 1800 36x 36 [10pt][ 1764 36x [10pt][ x 49 [10pt][ x 45 [10pt][ text{The speed of the train is 45 km/hr.} [10pt]]
Method 2: Simplified Calculation
Using a simplified approach, we can directly convert the speeds and distances into compatible units for quick calculations. Here, we consider the train's speed in m/s and the man's speed in km/hr. - Length of the train: 100 meters - Speed of the man: 5 km/hr (or 1.388889 m/s) - Time taken to cross the train: 7.2 seconds In 7.2 seconds, the man covers a distance of ( 7.2 times 1.388889 9.999998 ) meters, which is approximately 10 meters.
- Distance covered excluding the man's distance: ( 100 - 10 90 ) meters - Time taken: 7.2 seconds - Speed of the train: ( frac{90}{7.2} times frac{18}{5} 45 ) km/hr.Method 3: Relative Speed Formula
Using the concept of relative speed, where the relative speed is the sum of the speeds of the two objects when they are moving in opposite directions:
[ text{Relative speed} text{Train speed} text{Man speed} [10pt][ 100 text{ meters} div 7.2 text{ seconds} x times frac{5}{18} 1.388889 text{ m/s} [10pt][ 100 times frac{18}{7.2} 5x 18 times 1.388889 [10pt][ 100 times 2.5 5x 25 [10pt][ 250 5x 25 [10pt][ 250 - 25 5x [10pt][ 225 5x [10pt][ x 45 [10pt][ text{The speed of the train is 45 km/hr.} [10pt]]
Method 4: Detailed Speed Calculation
Another approach involves breaking down the speed calculation step-by-step:
- Relative speed: ( x 5 ) km/hr - Distance: 100 meters - Time: 7.2 seconds Converting the distance to kilometers and the time to hours, we get: [ text{Relative speed} frac{100 times 18}{7.2 times 1000} 50 text{ km/hr} [10pt][ 50 - 5 45 text{ km/hr} [10pt][ text{The speed of the train is 45 km/hr.} [10pt]]