Answer:
[tex]a_c = 571 m/s^2[/tex]
[tex]a_t = 30.77 m/s^2[/tex]
[tex]a = 571.83 m/s^2[/tex]
Explanation:
The angular speed of the ball when its tangential speed is 20m/s is:
[tex]\omega = v/r = 20 / 0.7 = 28.57 rad/s[/tex]
From here we can calculate the centripetal acceleration:
[tex]a_c = \omega^2r = 28.57^2*0.7 = 571 m/s^2[/tex]
The angular acceleration over t = 0.65s is:
[tex]\alpha = \omega/t = 28.57 / 0.65 = 44 rad/s^2[/tex]
So the tangential acceleration is
[tex]a_t = \alpha r = 44*0.7 = 30.77 m/s^2[/tex]
The magnitude of total acceleration, which is the combination of centripetal and tangential acceleration is:
[tex]a = \sqrt{a_t^2 + a_c^2} = \sqrt{30.77^2 + 571^2} = \sqrt{946.7929 + 326041} = \sqrt{326987.7929} = 571.83 m/s^2[/tex]
