Question

The 40th parallel of north latitude runs across the United States through Philadelphia, Indianapolis, and Denver. at this latitude earths radius is about 3030 miles. The earth rotates with an angular velocity of pi/12 radians per hour toward the east.

If a jet flies due west with the same angular velocity relative to the ground at the equinox the sun as viewed from the jet will stop in the sky. how fast in miles per hour with the jet have to travel west at the 40th parallel for this to happen?

Answer 1

Answer: 793 mi/h

Explanation:

We have the following data:

`R_(E)=3030 mi` is Earth radius at the 40th parallel of north latitude

`\omega_(E)=(\pi)/(12) rad/h` is the Earth's angular velocity (toward the east)

`\omega_(J)=(\pi)/(12) rad/h` is the Jet's angular velocity (toward due west)

And we need to find the Jet's speed
`V_(J)`, which is calculated by:

`V_(J)=\omega_(J)R_(E)`

`V_(J)=((\pi)/(12) rad/h)(3030 mi)`

`V_(J)=793.25 mi/h`