Answer:
[tex]\alpha =1.76\ rad/s^2[/tex]
Step-by-step explanation:
Given that,
Initial angular speed of the skater, [tex]\omega_i=4.5\pi\ rad/s[/tex]
Final angular speed of the skater, [tex]\omega_f=7.2\pi\ rad/s[/tex]
Time interval, t = 4.8 s
We need to find her average angular acceleration during this time interval. Let [tex]\alpha[/tex] be the average acceleration of the skater. So,
[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}\\\\\alpha =\dfrac{7.2\pi -4.5\pi}{4.8}\\\\\alpha =1.76\ rad/s^2[/tex]
So, her angular acceleration is [tex]1.76\ rad/s^2[/tex].