Answer:
[tex]a = 0.53 m/s^2[/tex]
Explanation:
initially the merry go round is at rest
after 6.73 s the merry go round will accelerates to 20 rpm
so final angular speed is given as
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 2\pi ( \frac{20}{60})[/tex]
[tex]\omega = 2.10 rad/s[/tex]
so final tangential speed is given as
[tex]v = r\omega[/tex]
[tex]v = 1.71 (2.10) = 3.58 m/s[/tex]
now average acceleration of the girl is given as
[tex]a = \frac{v_f - v_i}{\Delta t}[/tex]
[tex]a = \frac{3.58 - 0}{6.73}[/tex]
[tex]a = 0.53 m/s^2[/tex]
