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I've been studying special relativity and found a problem (at least for my limited knowledge): There's a car inside a train that is moving at speed $V_t$, for $V_t = 0.5c$. Once the car accelerates with $V_c = 0.5c$, from a reference frame other than those of the car and the train, does $V_t + V_c = c$?

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marked as duplicate by JamalS, ACuriousMind Nov 11 '16 at 15:45

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Assuming that $v_t$ and $v_c$ are measured with respect to the stationery observer on the ground.

For plain simple Galilean velocity addition one gets, $$v_{AC}=v_{AB}+v_{BC}\tag1$$ but using Einstein's velocity addition rule(with comes from using Lorentz transformations), its modified to, $$v_{AC}=\frac{v_{AB}+v_{BC}}{1+\frac{v_{AB}v_{BC}}{c^2}}\tag2$$

So using this knowledge, and having in mind that, $v_{CT}=0.5c$, $v_{TG}=0.5c$. Also I am assuming that the cars moves in the direction in which the train moves.

$(1)\implies v_{CG}=v_{CT}+v_{TG}=c$

$(2)\implies v_{CG}=\frac{v_{CT}+v_{TG}}{1+\frac{v_{CT}v_{TG}}{c^2}}=\frac{c}{1+0.5^2}=0.8c$

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  • $\begingroup$ I'm assuming that vCG will never reach c, then? $\endgroup$ – William Fernandes Nov 11 '16 at 13:04
  • $\begingroup$ An objects velocity cannot be greater than c. And if the velocity is c, its massless. $\endgroup$ – sbp Nov 11 '16 at 13:05

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