#### Answer

(a) $v = 3.9~m/s$
(b) The merry-go-round makes 2.5 revolutions.

#### Work Step by Step

(a) We can find the angular velocity as:
$\omega = \frac{2\pi~rad}{4.0~s}$
$\omega = 1.57~rad/s$
We can find the speed of a child on the rim as:
$v = \omega ~r$
$v = (1.57~rad/s)(2.5~m)$
$v = 3.9~m/s$
(b) We can find the angle $\theta$ that the merry-go-round rotates through as it slows down:
$\theta = (\frac{\omega_0+0}{2})(20~s)$
$\theta = (\frac{1.57~rad/s+0}{2})(20~s)$
$\theta = 15.7~rad$
We can convert $\theta$ to revolutions;
$\theta = (15.7~rad)(\frac{1~rev}{2\pi~rad})$
$\theta = 2.5~rev$
The merry-go-round makes 2.5 revolutions.