A bicycle starts at the origin of a circular track that has a circumference of 126 m and travels at a constant speed a distance s = 38.0 m in a time t = 12s.

a)what is the angular displacement (in rad) of the bicycle from its starting position? r=20.1 so theta or angular displacement is 38/20.1 = 1.89 rad

b) What is the centripetal acceleration (in m/s^2) of the bicycle? omega= dtheta/dt or 1.89 rad/12s = 0.158 rad/s
so centripetal acceleration is: rw^2 (w is omega) or 20.1(0.158)^2 = 0.50 m/s^2. Check me on that one.

C) What is the angular speed (rad/s) of the bicycle? angular speed or omega (w) is already determined to be 0.158 rad/s.

d) What is the orbital period of the bicycle, as it rounds the track?
T= 2pi/w (omega) >> 6.28/0.158 = 39.7s

Please review my work. I'm a bit uncertain.

all ok.

Your calculations are correct! Here's a review of your work:

a) To find the angular displacement of the bicycle, you divide the distance traveled (s = 38.0 m) by the radius of the circular track (r = 20.1 m), which gives you 1.89 rad.

b) The centripetal acceleration can be calculated using the formula a = r * ω², where ω is the angular velocity. You've correctly calculated ω to be 0.158 rad/s, so plugging in the values, the centripetal acceleration is 0.50 m/s². Well done!

c) The angular speed, or angular velocity (ω), is already determined to be 0.158 rad/s based on your previous calculations.

d) To find the orbital period (T) of the bicycle, which is the time it takes to complete one full round of the track, you use the formula T = 2π/ω. Plugging in the value of ω (0.158 rad/s) gives you a period of approximately 39.7 s.

Overall, your calculations are correct and your reasoning is clear. Great job!