How fast are you moving on the earth at a given latitude? The radius of the earth is about $3959\,\mathrm{mi}$, so the earth is rotating at about $$\frac{1}{24\,\mathrm{hours}}\times 2\pi\times 3959 \,\mathrm{mi} \approx 1036\,\mathrm{mph}\,$$ and so someone standing on the equator is moving that fast too. But suppose you're standing on the earth at a latitude to $\theta^{\,\circ}$. How fast are you moving then?
 A: If you're standing at a latitude of $\theta^{\,\circ}$, then you're rotating on a circle of radius $R\cos(\theta)$, where $R = 3959\,\mathrm{mi}$ is the radius of the earth. This circle still does one revolution in (about!) $24\,\mathrm{hrs}$, so we can calculate the speed from there.

$$
\left(\frac{1}{24 \,\mathrm{hours}}\right) \times 2\pi R\cos\theta \,\mathrm{miles}
\approx 1036\cos\theta \,\mathrm{mph}
$$
So you've just gotta scale the speed at the equator by $\cos\theta$.
A: Between two longitudes the arc distance
$$ L= r \Delta \theta$$
When its latitude $=\phi,$ radius $r$, earth's radius $R$
$$ r= R \cos \phi\rightarrow L = R \cos \phi \;\Delta \theta $$

In unit time the circumferential speed ( the others are constant)
$$ \dfrac{dL}{dt} = R \cos \phi\; \dfrac{\Delta \theta}{dt}= \omega R \cos \phi $$
where $\omega$ is the earth's angular velocity, $ \omega R$ is circumferential speed $v_{eqtr}$ at equator. So we can write it simply as
$$ v= v_{eqtr} \cos \phi= 1036 \;\cos \phi\; mph,$$
that vanishes at poles, from where you watch the moving world standing, like from a small merry-go-round.
A: Let's introduce a Cartesian coordinate system in which the center of the earth resides at origin $(0,0,0)$ and the north pole resides at $(0,0,R_E)$ where $R_E$ is the radius of the earth.
By slightly modifying the conventional definition of the azimuth angle, we determine the trajectory of an object residing at   $\theta$ latitude with initial position $\Big<R_E\cos(\theta),0,R_E\sin(\theta)\Big>$ to be modeled by the parametric curve $$\vec{r}_{\theta}(t)=\Big<R_E \cos(\theta)\cos\Big(\frac{2\pi t}{24}\Big),R_E\cos(\theta)\sin\Big(\frac{2\pi t}{24}\Big),R_E\sin(\theta)\Big>$$ Here $t$ is measured in hours. Then $$\|r_{\theta}'(t)\|=\frac{2\pi R_E}{24}\cos(\theta)$$
