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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?

Respuesta :

Answer:

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

Explanation:

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

[tex]\omega =\frac{v}{R}[/tex]

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

[tex]\frac{v_E}{2r}=\frac{v_R}{r}[/tex]

Now, solving for [tex]v_E[/tex] we get:

[tex]v_E=2v_R[/tex]

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

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