Question

A train of length $l=350m$ starts moving rectilinearly with constant acceleration $\omega =3.0\cdot {10}^{-2}m/s^2$; $t=30s$ after the start the locomotive headlight is switched on (event 1), and $\tau =60s$ after that event the tail signal light is switched on (event 2) . At what constant velocity $V$ (in m/s) relative to the Earth must a certain reference frame $K$ move for the two events to occur in it at the same point? (round off your answer to the nearest integer)

Answer 1

In the reference frame fixed to the train, the distance between the two events is obviously equal to $l$. Suppose the train starts moving at time $t=0$ in the positive x direction and take the origin ($x=0$) at the head-light of the train at $t=0$. Then the coordinate of first event in the earth's frame is
$x_1=\frac{1}{2}w{t}^2$
and similarly the coordinate of the second event is
$x_2=\frac{1}{2}w{(t+\tau )}^2-l$
The distance between the two events is obviously,
$x_1-x_2=l-w\tau (t+\frac{\tau }{2})=0.242km$
in the reference frame fixed on the earth.
For the two events to occur at the same point in the reference frame $K$, moving with constant velocity $V$ relative to the earth, the distance travelled by the frame in the time interval $T$ must be equal to the above distance.
Thus $V\tau =l-w\tau (t+\frac{\tau }{2})$
So, $V=\frac{1}{\tau }-w\tau (t+\frac{\tau }{2})=4.03m/s$
The frame $K$ must clearly be moving in a direction opposite to the train so that if (for example) the origin of the frame coincides with the point $x_1$ on the earth at time $t$, it coincides with the point $x_2$ at time $t+\tau$.