= Solution
An <Inhomogeneous Poisson process> with intensity $\lambda$ is a counting process with $N_0=0$, independent increments, and
$$
N_t-N_s\sim\operatorname{Poisson}(\Lambda(t)-\Lambda(s)),
\qquad
\Lambda(t)=\int_0^t\lambda(u)\,du.
$$
The continuity and positivity of $\lambda$ make $\Lambda$ strictly increasing.
For $M_t=N_{g(t)}$ and $0\leq s<t$,
$$
M_t-M_s=N_{g(t)}-N_{g(s)}
$$
has a Poisson distribution with mean
$$
\Lambda(g(t))-\Lambda(g(s))=t-s.
$$
Disjoint increments remain independent, so $M$ is a homogeneous rate-one <Poisson process>. Taking $g=\Lambda^{-1}$ gives $N_t=M_{\Lambda(t)}$, and hence
$$
\boxed{\ N_t\sim\operatorname{Poisson}(\Lambda(t)).\ }
$$
This is the <time change of an inhomogeneous Poisson process>.
For the bicycle model, let an arrival occur at time $s\leq t$ and have velocity $V$. At time $t$ its position is $V(t-s)$, so it lies in the first $x$ miles exactly when
$$
V(t-s)\leq x.
$$
Conditional on its arrival time, independently marking the bicycle by whether this event occurs is <Poisson thinning>. Therefore the desired count is Poisson with mean
$$
\begin{aligned}
\lambda\int_0^t\mathbb P(V(t-s)\leq x)\,ds
&=\lambda\mathbb E\int_0^t\mathbf1_{\{Vu\leq x\}}\,du\\
&=\lambda\mathbb E\left[\min\left(t,\frac{x}{V}\right)\right]\\
&=\lambda\mathbb E\left[V^{-1}\min\{x,Vt\}\right].
\end{aligned}
$$
Thus
$$
\boxed{\ \#\{\text{bicycles in the first }x\text{ miles at time }t\}
\sim\operatorname{Poisson}\!\left(\lambda\mathbb E[V^{-1}\min\{x,Vt\}]\right).\ }
$$
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