Question

There is a freight train which is 100 meters long in its own reference frame. It passes through a railway platform that has a length of 80 meters in its own reference frame. The train travels past the platform so fast that, in the platform's reference frame, both the train and platform are the same length. How fast must the train be moving?

Penalty factor 25%
a. 0.2c cross out
b. 0.4c cross out
c. 0.6c cross out
d. 0.8c cross out
e. none of the other answers is correct cross out

Answer 1

The train must be moving at approximately 0.6 times the speed of light to appear the same length as the 80-meter platform from the platform's reference frame, due to the relativistic effect of length contraction.

Step-by-step explanation:

The question is asking about the relativistic effects on length when an object is moving at a significant fraction of the speed of light (symbolized as c). According to the principles of special relativity, the measured length of an object in motion—its length contraction—will be shorter in the frame of an observer that views the object moving relative to them.

To find the speed at which both the train and the platform appear to be of the same length, we use the Lorentz factor (γ), which relates to the speed of the train (v) as follows: γ = 1 / sqrt(1-(v^2/c^2)) where c is the speed of light. Since the train's rest length is 100 meters and it appears the same length as the 80-meter platform in the platform's frame, we can say that the contracted length of the train (L') is 80 meters.

Thus, L' = L / γ where L is the rest length of the train. Plugging in the values and rearranging for v, we find that v ≈ 0.6c. This means the train must be moving at roughly 0.6 times the speed of light for both the train and platform to appear as the same length to an observer on the platform.