A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the cyclist begins to uniformly apply the brakes. The bicycle stops in 5.0 s.
-How far did the bicycle travel during the 5.0 seconds of braking?
A)1.8 m
B)8.8 m
C)22 m
D)42 m
E)44 m

Question

Quizplus · Accepted Answer

First, we need to find the initial linear speed of the bicycle. Since it is rolling without slipping, the linear speed is equal to the product of the angular speed and the radius:
v = ωr = (2.0 rev/s)(2π)(0.70 m) = 8.8 m/s

During the 5.0 s of braking, the bicycle experiences a constant deceleration that causes it to stop. The deceleration can be found using the kinematic equation:
Δx = v0t + ½at^2

where Δx is the distance traveled during the braking, v0 is the initial speed, a is the deceleration, and t is the time.

Since the bicycle stops, its final speed is zero, so we have:
0 = v0 - at
a = v0/t = 8.8 m/s / 5.0 s = 1.8 m/s^2

Substituting this into the kinematic equation, we get:
Δx = v0t + ½at^2 = (8.8 m/s)(5.0 s) + ½(1.8 m/s^2)(5.0 s)^2 = 22 m

Therefore, the bicycle traveled a distance of 22 m during the 5.0 s of braking. Choice C is the best answer.

A Bicycle Wheel of Radius 0