Answer
(a) The linear speed is 1.9 m/s
(b) The centripetal acceleration is $3.0~m/s^2$.
Since we can assume that the merry-go-round is rotating at a constant speed, the tangential acceleration is zero.
Work Step by Step
(a) We can calculate the linear speed $v$ by:
$v = \frac{distance}{t} = \frac{2\pi r}{t} = \frac{2\pi (1.2~m)}{4.0~s} = 1.9~m/s$
The linear speed, therefore, is 1.9 m/s.
(b) We can calculate the centripetal acceleration $a_c$ as:
$a_c = \frac{v^2}{r} = \frac{4\pi^2 ~r}{t^2}$
$a_c = \frac{(4 \pi^2) (1.2~m)}{(4.0~s)^2} = 3.0~m/s^2$
The centripetal acceleration is $3.0~m/s^2$.
Since we can assume that the merry-go-round is rotating at a constant speed, the tangential acceleration is zero.
