A ball with radius .15 m is rolling with an angular velocity of 6.5 rad/s. (a) What linear distance will the ball roll in 5.0 seconds? The ball then slows down with a linear acceleration of -1.2 rad/s2. (B) How fast will it be rolling rad/s after .65 second? (C) If a piece of gum is stuck on the outer part of the ball, what is its linear speed? (4.9, 5.72, .858)

(a) 4.875 m

(b) 5.72 rad/s

(c) 0.858 m/s

Explanation:

(a) Assuming constant angular speed, the angular distance the ball would have traveled after 5s at the rate of 6.5 rad/s is

6.5 * 5 = 32.5 rad

With radius of 0.15m, the linear distance it would have traveled is

32.5 * 0.15 = 4.875 m

(b)The angular velocity of the ball after 0.65s when subjected to an angular acceleration of -1.2 rad/s is

6.5 - 1.2*0.65 = 5.72 rad/s

(c)The linear speed of the ball is the product of the angular speed and radius 0.15 m

5.72 * 0.15 = 0.858 m/s

b.

explanation:

10.0

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000