# iit-jee 2008 relativity problem

## iit-jee 2008 relativity problem

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?

**Pinnaka**- Admin
- Posts : 36

Join date : 2008-05-20

Age : 31

Location : India

## Re: iit-jee 2008 relativity problem

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

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

**Pinnaka**- Admin
- Posts : 36

Join date : 2008-05-20

Age : 31

Location : India

Similar topics

» MOA-2008-BLG-310Lb - planet in the Galactic Bulge

» Silverfish problem....

» OGLE-2008-BLG-513L b

» Pls explain- National Budget Circular 517 dated dec 22, 2008.

» Mercredi 5 novembre 2008: Un jour historique!!!

» Silverfish problem....

» OGLE-2008-BLG-513L b

» Pls explain- National Budget Circular 517 dated dec 22, 2008.

» Mercredi 5 novembre 2008: Un jour historique!!!

Page

**1**of**1****Permissions in this forum:**

**cannot**reply to topics in this forum