= Solution
Align both frames' $x$ axes with the motion and make their origins coincide at $t=t'=0$. With $S'$ moving along $+x$ at <speed> $v_2$ relative to $S$, the <Lorentz transformation> and its inverse are
$$
x'=\gamma_2(x-v_2t),\quad t'=\gamma_2\left(t-\frac{v_2x}{c^2}\right),\qquad
x=\gamma_2(x'+v_2t'),\quad t=\gamma_2\left(t'+\frac{v_2x'}{c^2}\right),
$$
where $\gamma_2=(1-v_2^2/c^2)^{-1/2}$; the transverse coordinates are unchanged. Along the object's path, $dx'/dt'=v_1$. Taking differentials of the inverse transformation gives
$$
\frac{dx}{dt}=\frac{\gamma_2(dx'+v_2dt')}{\gamma_2(dt'+v_2dx'/c^2)},
\qquad
\boxed{v=\frac{v_1+v_2}{1+v_1v_2/c^2}.}
$$
This is <relativistic velocity addition>. For $v_1,v_2\ll c$ it reduces to the Galilean sum; if $v_1=c$, the result is still $c$. Two subluminal positive <speeds> give a subluminal result.
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