An automobile traveling 80 mi/hr has wheels 31 inches in diameter. If the car is brought to a stop uniformly in 58 turns of the wheels, what is the magnitude of the angular acceleration of the wheels (in radians/second^2)?

Answer 1

C = pi * 2r = 3.14 * 31 = 97.4 In=8.1Ft.

d = 58rev * 6.28rad/rev = 364.24 Rad.
V=80mi/h*5280Ft/mi*6.28rad/8.1Ft*1h/3600s = 91 rad/s. = Vo.

V^2 = Vo^2 + 2a*d
a = (V^2-Vo^2)/2d
a = (0-(91)^2)/728.48 = -11.4 rad/s^2