The velocity of a spinning gyroscope drops from 12 rad/s to 6 rad/s due to deceleration of - 1.2 rad/s^2. how much time expires during this process? how many radians does the top spin during the process?

5 seconds expired during the deceleration. Top rotated 45 radians during these 5 seconds. First, calculate the chance in velocity, by subtracting the initial velocity from the final velocity. So 6 rad/s - 12 rad/s = - 6 rad/s So we lost a total of 6 rad/s. Divide that by the deceleration to give the number of seconds. So - 6 rad/s / - 1.2 rad/s^2 = 5 s So it takes 5 seconds for the deceleration to happen. The equation that expresses the number of radians performed under constant deceleration with an initial velocity is d = VT + 0.5 AT^2 where d = distance V = initial velocity T = time A = acceleration Substituting the known values gives. d = VT + 0.5 AT^2 d = 12 rad/s * 5s + 0.5 * - 1.2 rad/s^2 (5s) ^2 d = 60 rad - 0.6 rad/s^2 * 25s^2 d = 60 rad - 15 rad d = 45 rad So the top rotated 45 radians while decelerating from 12 rad/s to 6 rad/s