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Galilean Transformations

  • Thread starter austim14
  • Start date
  • #1
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βI solved this problem but I do not know if it is correct becasue there is no way to check it:
Imagine that we define the rear end of a train 120 m long to long to define the origin X'=0 in the train frame and we define a certain track signal light to define the origin X=0 in the track frame. Imagine that the rear end of of the train passes this sign at t+t'=0 as the train moves in the +X direction at a constant speed of 25 m/s. 12 seconds later, the engineer turns on the train's heasdlight. A) Where does this event occur in the train frame? B) Where does this event occur in the track frame



Homework Equations



Use Galilean transformation equations for position. β is the contstant velocity.
t=t'
x'=X- β*t


The Attempt at a Solution


Here is my work: 25m/s*12s=300m
300m+120m= 420m
Train Frame 420m

420m=X-300m
X=720m
Track frame 720m


Is this correct?
 
Last edited:

Answers and Replies

  • #2
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Here is my work: 25m/s*12s=300m
300m+120m= 420m
Train Frame 420m
What is the speed of the train relative to the train (=in the train frame)?

I agree with the calculations of the second part if you fix the value from (A).
 
  • #3
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I think it's the 25 m/s
 
  • #4
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If the train is 120 M long, and the rear of the train is at x' =0, what is the x' coordinate of the front of the train? (presumably the engineer is situated in the front of the train). (Assume that you have a set of coordinates permanently laid out along the floor of the train from the rear of the train to the front of the train). The front of the train is always situated at this x' location.

Assume you also have a set of coordinates permanently laid out along the side of the track starting at the signal at x = 0. At time t = t' = 0, what are the x coordinates of the rear and the front of the train? During the 12 seconds, how far does the front of the train advance along the x coordinates at the side of the track? What is its x coordinate after these 12 seconds?
 

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