Question: A train of length $l=350 metres$ starts moving rectilinearly with the constant acceleration $a=3×10^-² ms^-²$; $t=30s$ after the start the locomotive headlight is switched on $(event 1)$, and $t=60s$ after that event the tail signal light is switched on $(event 2)$. How and 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$?$

Is this even possible? I could not think how to start. I thought of relativistic mechanics but I am unsure if it will help either. I thought of replacing $K$ frame with aeroplane to visualise better. Please help and thanks in advance$!$ Answer is 'towards train with velocity $V=4 ms^-¹$'.


1 Answer 1


The train is not going fast enough for relativistic effects to be significant.

If we take the front of the train as being at $x=0$ at $t=0$ then after $30$ seconds the front of the train has moved a distance

$\displaystyle \frac 1 2 a t^2 = 1.5 \times 10^{-2} \times 900 = 13.5$ metres

so the $(x,t)$ co-ordinates of event $1$ (relative to the earth reference frame) are $(13.5, 30)$.

After a further $60$ seconds, at $t=90$, the train has moved

$1.5 \times 10^{-2} \times 8100 = 121.5$ metres

but the back of the train is $350$ metres behind the front, so the the $(x,t)$ co-ordinates of event $2$ are $(-228.5, 90)$.

If reference frame $K$ is moving at velocity $v$ relative to earth then an event with co-ordinates $(x,t)$ relative to earth has co-ordinates $(x',t)$ relative to $K$, where

$x' = x - vt$

If event $1$ and event $2$ occur at the same point in frame $K$ then their $x'$ co-ordinates are the same so we have

$13.5 - 30v = -228.5 - 90v$

and from this you can find $v$.


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