Answer:
the acceleration of the small bug is 12.83 m/s²
Step-by-step explanation:
Given;
time of motion, t = 0.5 s
radius of the circular path created by his arm, r = 1.3 m
if he rotates his arm from horizontal to vertical, the angular displacement = 90⁰
The centripetal acceleration of the ball is calculated as;
![a_c = \omega^2 r\\\\a_c = ((\theta)/(t) )^2 r\\\\](https://img.qammunity.org/2022/formulas/physics/college/mswriwmnh8quoraz9b27rcpg7upr6uyatt.png)
![a_c = ((90)/(360) *( 2\pi )/(t) )^2r\\\\a_c = ((\pi)/(2t) )^2 r\\\\a_c = (\pi^2r)/(4t^2) = (\pi^2 *1.3 )/(4* 0.5^2) = 12.83 \ m/s^2](https://img.qammunity.org/2022/formulas/physics/college/3ew32zxvvkpl9cn26ch9ja0wynx0j8tsy3.png)
Therefore, the acceleration of the small bug is 12.83 m/s²