Emily rides on a horizontal rotating platform of radius r at an amusement park and moves at a linear velocity, vt = 1 m/s, one third the way from the center to the
outer edge, and a rotational velocity of 2 RPM. How much would her linear and
rotational velocities be if she moves to the outer edge?

Her linear velocity on the outer edge is 3 m/s

Her rotational velocity, is constant at 2 RPM

Explanation:

The given parameters are;

The radius of the rotating platform = r

The linear velocity  v(t) = 1 m/s

The position Emily is riding = 1/3 the distance from the center

The rotational velocity = 2 RPM = 2 Revolutions per minute

The formula for angular velocity, ω is given as follows;

ω = θ/t

Where;

θ = The angle rotated

t = The time taken for the rotation

Given that the rotational velocity = 2 RPM

1 revolution = 2·π radians

The angular velocity, ω = 2×π×(2 RPM) = 4·π rad/min = 4·π/60 rad/seconds

ω = 4·π/60 rad/seconds

The linear velocity, v = r₁×θ/t = r₁×ω

Where;

r₁ = 1/3 × r

v = 1 m/s = 1/3 × r  × 4·π/60 rad/seconds

∴ r = 1/(1/3 × 4·π/60) = 45/π meters

r = 45/π meters

Therefore her linear velocity, v₂ and her rotational velocity if she moves to the outer edge will be given as follows;

v₂ = r × ω = 45/π × 4·π/60 = 3 m/s

Her linear velocity on the outer edge, v₂ = 3 m/s

Her rotational velocity, remains at 2 RPM.

wavelength divides in half and speed stays the same.

i think its c if i'm wrong