= Constant-burn rocket with ground friction
{title2=$q(t)\dot v=su-\mu gq(t)$}
For fuel mass rate $s>0$ and backward relative exhaust speed $u$, a rocket sliding on a horizontal plane has remaining mass $q(t)$ and satisfies the displayed <rocket equation> with <kinetic friction>. Integration over a sliding interval gives $v(t)-v(t_0)=u\log[q(t_0)/q(t)]-\mu g(t-t_0)$. From initial rest, immediate forward sliding additionally requires thrust to overcome <static friction>. If both friction coefficients equal $\mu$, motion starts only when $su>\mu gq(t)$. An algebraically negative sliding speed is not the physical solution at rest.
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