= Solution
Use the <prescribed-speed time parametrization of a curve>. On a regular part traversed in increasing $u$, let $q(u)=\|F'(u)\|>0$ and assume the prescribed <speed> $v(F(u))$ is positive. The chain rule for <velocity> gives
$$
\left\|{dF\over dt}\right\|=q(u){du\over dt}=v(F(u)),\qquad \boxed{{du\over dt}={v(F(u))\over q(u)}}.
$$
Integrate this scalar <ordinary differential equation> with an adaptive <Runge-Kutta method>, arranging output at the exact times $t_k=t_0+k\Delta t$. Evaluate $P_k=F(u(t_k))$. Internal integration steps may be shorter than $\Delta t$; output interpolation must meet the same error tolerance, so variable numerical steps do not create variable frame times.
An equivalent implementation precomputes
$$
t(u)-t(0)=\int_0^u{\|F'(\xi)\|\over v(F(\xi))}\,d\xi
$$
by adaptive <numerical integration>. This function is strictly increasing. Bracket each $t_k$ in the table and invert using safeguarded <Newton method> or <interval bisection>; monotonicity makes each solution unique. Stop at $u=1$, with a shorter final time interval if the total travel time is not a multiple of $\Delta t$. This accounts for changes in both the curve's parameter <speed> and the physical <speed>, unlike equal increments in $u$.
If the specified <speed> vanishes, examine the reciprocal-speed integral at that event. It may diverge, giving asymptotic approach, or be integrable, giving arrival in finite time. A scalar <speed> does not determine whether motion reverses or how it resumes after stopping; such behavior needs the physical dynamics or a chosen continuation. The positive-speed formula applies between these events, with the direction changed if the traversal reverses.
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