Answer:
The linear spin speed would be approximately [tex]850\; {\rm mph}[/tex] at [tex]35^{\circ}\; {\rm N}[/tex] or [tex]35^{\circ}\; {\rm S}[/tex], assuming that the Earth is a perfect sphere.
Explanation:
Regardless of the latitude on the planet, the angular velocity [tex]\omega[/tex] of the spin of the Earth would be the same. The "speed" in this question refer to the linear speed [tex]v[/tex] at the given position. If [tex]\omega[/tex] stays the same, the value of [tex]v[/tex] would be proportional to the distance [tex]r[/tex] from the axis of rotation:
[tex]v = \omega\, r[/tex].
Let [tex]r_{E}[/tex] denote the average radius of the Earth. Under the assumption that the Earth is a perfect sphere, the distance between a point on the equator of the Earth and the axis of rotation would also be [tex]r_{E}[/tex].
For a point at a latitude of [tex]\phi[/tex] ([tex]{\rm N}[/tex] or [tex]{\rm S}[/tex]) on the surface of the planet, the distance from the axis of rotation would be:
[tex]r = r_{E}\, \cos\left(\phi\right)[/tex].
In other words, the distance between the axis of rotation and a point at a latitude of [tex]\phi[/tex] ([tex]{\rm N}[/tex] or [tex]{\rm S}[/tex]) on the Earth is proportional to [tex]\cos\left(\phi\right)[/tex]. Because angular velocity of spin is the same across the planet, linear velocity of the spin would be proportional to this distance, and would thus be proportional to the cosine of the latitude angle [tex]\cos(\phi)[/tex].
In this question, at [tex]35^{\circ}\; {\rm N}[/tex] or [tex]35^{\circ}\; {\rm S}[/tex], this distance from the axis at the center of rotation would be [tex]r_{E}\, \cos\left(35^{\circ}\right)[/tex]. Given that the linear speed of rotation is [tex]1\, 037\; {\text{mph}}[/tex] at the equator where the distance is [tex]r_{E}[/tex], the speed at [tex]35^{\circ}\; {\rm N}[/tex] or [tex]35^{\circ}\; {\rm S}[/tex] would be:
[tex]\begin{aligned} & (1\, 037\; \text{mph})\times \frac{r_{E}\, \cos\left(35^{\circ}\right)}{r_{E}} \\ =\; & (1\, 037\; \text{mph})\, \cos\left(35^{\circ}\right) \\\approx\; & 850\; \text{mph}\end{aligned}[/tex].