July 29, 2008

Q. 1.14 - "Problems in General Physics" by I.E. Irodov

Question: A train of length$l$= 350 m starts moving rectilinearly with constant
acceleration$w$=$3.0 \times 10^{-2}$;$t$= 3 s after the start the locomotive headlight
is switched on (event$1$), and$\tau$= 60 s after that event the tail signal light is switched
on (event$2$). Find the distance between these event in the reference frames fixed to the train
and to the earth. How and at what constant velocity$V$relative to the Earth must a certain
reference frame$K$move for the two events to occur in it at the same point.

Answer:
In the reference frame fixed to the train, obviously the distance between the two points is$l$=
350 m.

For the reference frame fixed to earth, lets fix the initial position of headlight as the origin and
the direction of train movement as given by unit-vector$\vec{i}.$

In this reference frame, position of headlight at time$t$is$(\frac{1}{2}wt^2)\vec{i}.$This is
the position of event$1$.

After$\tau + t$, the position of headlight is$(\frac{1}{2}w(\tau+t)^2)\vec{i}.$

Thus, the position of taillight at time$\tau + t$is$(\frac{1}{2}w(\tau+t)^2 - l)\vec{i}.$This is
the position of event$2$.

Thus, distance between these two events:

\begin{eqnarray*}<br />& = & \frac{1}{2}(w(\tau+t)^2 - t^2) - l \\<br />& = & \frac{1}{2}...<br />...& \frac{1}{2}w(\tau+2t)\tau - l \\<br />& = & w\tau (t + \tau/2) - l<br />\end{eqnarray*}


Posted by Apurva Sharan at 2:43 PM
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