After t seconds the train headlight has moved
and 
later it will be 
. The taillight however, will be l units behind the head light so event 2 will occur at
. The distance between the two events thus events are separated by,
.
If a frame is moving at speed V then, during time
it will move a distance of V. This distance must exactly be equal to the separation of distance between the two events. Thus, the velocity of the frame should simply be,
6 comments:
i did not understand the soln
dear siva if you can tell me at which point in the solution you failed to understand I can try and answer you better
sir,
gd evenin, i dint undersand why u hav divided the equation by time 2.i think we hav to mltiply it na because we want to find out distance between those 2 events w.r.t reference frame train itself..
All calculations in the problem are w.r.t a stationary frame of reference. Maybe I should draw a figure to explain the solution... let me work on it and try to make it clearer.
ok I've added a small figure and changed the last equation to make things a bit more clearer. Hope this helps.
well done , i never thought that the solution of this question will be this simple , well i am nt so good at phy. , but this solution made me proud , i am happy to see this answer :)
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