A marching band consists of rows of musicians walking in straight, even lines. When a marching band performs in an event, such as a parade, and must round a curve in the road, the musician on the outside of the curve must walk around the curve in the same amount of time as the musician on the inside of the curve. This motion can be approximated by a disk rotating at a constant rate about an axis perpendicular to its plane. In this case, the axis of rotation is at the inside of the curve. Consider two musicians, Alf and Beth. Beth is four times the distance from the inside of the curve as Alf. Knowing that If Beth travels a distance s during time Δt, how far does Alf travel during the same amount of time= (1/4)s. If Alf moves with speed v, what is Beth's speed?

a) Alf travels a quarter of the distance of Beth

b) linear velocity of Beth is 4 times greater than the linear velocity of Alf

Explanation:

This is an exercise that we can solve using the angular kinematic relations and its relation to the linear scientific one

             s = θ r

             v = ω r

the distance relationship between the two musicians is

if alf is at a distance r₁ = R₀ from the center of the curve

Beth has a distance r₂ = 4 R₀

a) ask for the distance traveled by Alf

Let's find the angle that Beth travels (Musician 2)

            θ = s₂ / r₂

the angle traveled is the same for the two musician, the angle for Alf is

            θ = s₁ / r₁

             s₂ /4R₀ = s₁ /R₀

             s₁ = s₂ / 4

therefore Alf travels a quarter of the distance of Beth

b) Beth's linear velocity

look for the angular velocity that is the same for the two musicians

           w = v₁ / r₁

            w = v₂ / r₂

we equalize

           v₁ / r₁ = v₂ / r₂

           v₁/R₀ = v₂/4R₀

            v₂ = 4 v₁

therefore the linear velocity of Beth is 4 times greater than the linear velocity of Alf

i think its d if wrong im sorry