Solving Train Velocity Problems Using Relative Motion and Simple Algebra

Understanding the Problem: Relative motion and the algebraic calculation can be used to solve problems involving the velocity of trains or other moving objects. This article provides a detailed explanation of such a problem, including a step-by-step solution. Additionally, it offers an alternative solution using relative velocity and simple algebraic equations, making it accessible and easy to follow for students and enthusiasts alike.

Consider this scenario: A passenger on a train traveling at a speed of 40 km/h observes that a train 75 meters long passes by the window in 3 seconds. What is the speed of the oncoming train?

Approach 1: Using Distance Speed × Time

Let's start by understanding the problem using the basic algebraic equation for distance:

Distance Speed × Time

The length of the oncoming train 75 meters

The time taken for the train to pass the window 3 seconds

Convert the speed of the first train to meters per second (m/s): 40 km/hr 40000 m/hr 40000 / 3600 m/s 11.11 m/s

Using the formula: Distance Speed × Time

75 meters (40000 / 3600) m/s × 3 seconds

75 meters 11.11 m/s × 3 seconds

75 meters 33.33 meters per second × 3 seconds

75 meters 99.99 meters (approx 100 meters)

Now, calculate the relative speed of the two trains (the oncoming train the first train's speed):

11.11 m/s speed of the oncoming train 50 m/s (relative speed)

speed of the oncoming train 50 m/s - 11.11 m/s 63.89 m/s

The relative speed of 63.89 m/s can be converted back to km/h:

63.89 m/s 63.89 × (3600 / 1000) km/h 229.9 km/h (approx 50 km/h)

Approach 2: Using Relative Velocity

Alternatively, we can use the concept of relative velocity. The passenger observed that the oncoming train passed by the window in 3 seconds. This means the total length of 75 meters was covered at a relative velocity.

Given:

v_1 40 km/h 11.11 m/s (speed of the first train)

Relative velocity (v) between the two trains is 11.11 m/s (since they are moving towards each other)

Time (t) taken to pass the window 3 seconds

Total length (s) 75 meters

Using the relative velocity formula:

s v × t

75 meters (11.11 m/s v) × 3 seconds

75 meters 33.33 m/s 3v

Solving for v:

75 - 33.33 3v

41.67 m/s 3v

v 13.89 m/s

The speed of the oncoming train 13.89 m/s 13.89 × (3600 / 1000) km/h 50 km/h

Conclusion:

By using the basic algebraic approach or the relative velocity method, we can determine the speed of the oncoming train. In this case, it is 50 km/h. This problem is a practical application of the principles of relative motion and is useful in various real-world scenarios, such as traffic analysis or engineering.

Keywords:

Train Velocity: The speed at which a train is traveling. Velocity is a vector quantity that includes speed and direction.

Relative Motion: The motion of an object in relation to another object or a fixed reference point.

Algebraic Problem Solving: Using equations and algebra to solve for unknown quantities in a problem.