Question

Ethan and Rebecca are riding on a merry-go-round. Ethan rides on a horse at the outer rim of the circlar platform, twice as far from the center of the circular platform as Rebecca, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, describe Ethans tangential speed?

Answer 1

Ethan's tangential speed would be twice that of Rebecca's on the merry-go-round since he is twice as far from the center and the merry-go-round rotates at a constant angular speed.

Step-by-step explanation:

When Ethan and Rebecca are riding on a merry-go-round and Ethan is twice as far from the center as Rebecca, Ethan's tangential speed will be twice that of Rebecca's if the merry-go-round rotates at a constant angular speed. This is because tangential speed (v) is directly proportional to the radius (r) from the center for a given angular speed (ω). Since Ethan is at a distance from the center that is twice the radius of Rebecca's position, his tangential speed will be twice as much, as represented by the equation v = rω. Therefore, if Rebecca has a tangential speed v, Ethan will have a tangential speed of 2v.

Answer 2

Ethan's tangential speed is twice as Rebecca's.

Step-by-step explanation:

Let
`r` be the distance from the center of the platform to Rebecca's horse,
`v_R` Rebecca's tangential speed,
`v_E` the Ethan's tangential speed and
`\omega` the merry-go-round angular speed. The distance from the center of the platform to Ethan's horse is
`2r`. The angular speed is related to the tangential speed by the equation:

`\omega =(v)/(R)`

Since the angular speed is the same for Ethan and Rebecca, we have that:

`(v_E)/(2r)=(v_R)/(r)`

Now, solving for
`v_E` we get:

`v_E=2v_R`

It means that Ethan's tangential speed is twice as Rebecca's.