# Relativistic or non relativiistic mechanics?

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 $$V=4 ms^-¹$$'.

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 $$v$$.