iit-jee 2008 relativity problem

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iit-jee 2008 relativity problem

Post  Pinnaka on Wed May 21, 2008 12:52 am

Investigate following phenomenon: while in a moving train looking out of window, the objects at horizons do not move too fast. But objects close to window (e.g. telegraph poles) move extremely fast behind the window. How is this apparent speed related to the distance from the window?

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Re: iit-jee 2008 relativity problem

Post  Pinnaka on Wed May 21, 2008 12:54 am

Assuming that the train is moving constant speed V. the calculations are as follows
Let the distance between the window and the object being observed be d.
Now the relative velocity between the observer in the train and the point being observed is the rate of change of the segment connecting these two points. Therefore being still at a particular point and remaining still there is no relative velocity. lets us take the path of the train relative to the object being observed.



the side labeled x represents the motion of the train or the observer. the vertex of the triangle (right) represents the point being observed. Therefore the relative velocity between the observer and the point being observed is rate of change of the side labeled l.

right triangle
l^2 = d^2 + x^2
differentiating both sides
2l dl/dt = 2x dx/dt ( d is constant)
dl/dt = x*V/l where V =dx/dt
dl/dt gives the relative speed required

V_relative = V^2t/{sqrt(d^2 + v^2*t^2)}
Therefore the relative velocity is inversely proportional to the sqrt(d^2 + v^2*t^2)}
This equation makes sense
Taking the limit of the above expression as x goes to infinity
the limit turns out to be proportional to 1/{sqrt(d^2+v^2*t^2 ) as it goes to infinity
so the velocity is zero
it makes sense



Please do visulaise a triangle and draw a figure

note: this question has been answered by expert pinnaka

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