The earth's radius is 6.37×106m; it rotates once every 24 hours.
What is the speed of a point on the earth's surface located at 1/4 of the length of the arc between the equator and the pole, measured from the equator? (Hint: what is the radius of the circle in which the point moves?)

Speed of the point = 1.541 × 10⁶ m/h = 428 m/s

Explanation:

Modelling the earth as a sphere,

The equator and the pole have a length of an arc that's fully 90° of the whole 360° circular circumference.

A point located 1/4 of the length of the arc between the equator and the pole has an angle (1/4) × 90° arc = 22.5°.

On the attached image, θ = 22.5°, so Φ = 90° - 22.5° = 67.5°

To obtain the radius of the point on the globe

From the image again, it shows that trigonometric relations can be used to obtain this radius

Sin Φ = (radius of the point)/(radius of the earth)

Sin 67.5 = radius of the point/(6.37 × 10⁶)

Radius of the point = 6.37 × 10⁶ × 0.9239 = 5.885 × 10⁶ m

Speed of the point on the globe = (circumference of the circular path mapped out by the point during rotation)/(24 hours)

circumference of the circular path mapped out by the point during rotation = 2πr = 2π × 5.885 × 10⁶ = 3.6977 × 10⁷ m

Speed of the point = 3.6977 × 10⁷/24 = 1.541 × 10⁶ m/h = 428 m/s

i think the answer would be b