9:37 PM Wed Sep 14 . . . 75% 4 Review | Constants SOLUTION SET UP (Figure 1) shows our diagram. In this example the motion is in a vertical circle. Passengers in a carnival ride travel in a circle with SOLVE We will use the equation a = 2 / R. To find the speed v, we use the fact that a passenger travels a distance equal to the radius 5.0 m. The ride moves at a constant speed circumference of the circle (27 R) in the time T for one revolution: and makes one complete circle in a time T = 4.0 s. What is the acceleration of the passengers? U = 2TR 27 (5.0 m) 4.0 s = 7.9m/s The centripetal acceleration is (7.9 m/ s) 2 arad = D 5.0 m = 12 m/$2 REFLECT The direction of a is toward the center of the circle. The magnitude of a is greater than g, the acceleration due to gravity, so this is not a ride for the faint-hearted. (But some roller coasters subject their passengers to accelerations as great as 4g.) Part A - Practice Problem: If the ride increases in speed so that T' = 3.5 s, what is arad ? (This question can be answered by using proportional reasoning, igure without much arithmetic.) Express your answer in meters per second squared. T - 4.0 s IVO AED ? R - 5.0 m m/s2 Submit Request