how much faster does a climber on top of the mountain move than a surfer at a nearby beach? the earth's radius is 6400 km .

When the object is oscillating about the origin, repeatedly going left and right by identical amounts it has a time period T where the period T is the time required for one complete oscillation or cycle. The angular frequency ω is related to the period of the motion by an equation in the form

ω[tex]=\frac{2\pi }{T}[/tex]

Where T is the period of the earth. The period of the earth to complete one cycle is 24 hours, so it is in seconds given by

T = [tex]( 24 h)\frac{3600 s}{ 1 h}[/tex] = 86.4 × 10³ s

Now, we plug the value for T into equation (1) to get ω into rad/s

ω [tex]=\frac{2\pi }{T} \\[/tex]

ω = [tex]\frac{2\pi }{86.4 * 10^{3} s}[/tex]

ω = 7.27 × [tex]10^{-5}[/tex] rad/s

Both persons at the beach and at the top of the mountain have a velocity that is tangent to the circle of motion, and its acceleration is called centripetal acceleration. Any object on earth has angular velocity ω and a tangential speed v and they are related to each other by an equation in the form

v = ω r

Where r is the radius of the circle of the motion.

For the person at the beach, its radius of motion is the same as the radius of the earth r = R while for the person at the top of the mountain its radius of motion is [tex]R_{E}[/tex] + 3000

[tex]v_{Beach}[/tex] = ω[tex]R_{E}[/tex]

= [tex]( 7. 27 * 10^{-5} rad/s ) (6400 * 10^{3} )[/tex]

= 465.280 m/s

[tex]v_{Mountain}[/tex] = ω[tex](R_{E} + 3000)[/tex]

= [tex]( 7.27 * 10^{-5} rad/s) ( 6400 * 10^{3} + 3000m )[/tex]

= 465.498 m/s

As shown by the results, the person at the top of the mountain has a fast speed. The difference between both velocities is expressed by

Δv =[tex]v_{Mountain}[/tex] - [tex]v_{Beach}[/tex]

= ( 468.498 - 465.280) m/s

= 0.22 m/s

= 22 cm/s

Hence, The person at the top of the mountain is faster than the person at the beach by 22cm/s.