Question

The blade of a windshield wiper moves through an angle of 90.08 in 0.40 s. The tip of the blade moves on the arc of a circle that has a radius of 0.45 m. What is the magnitude of the centripetal acceleration of the tip of the blade

Answer 1

The centripetal acceleration is 6.95 m/s²

Step-by-step explanation:

Given;

angular displacement of the blade, θ = 90.08⁰

duration of motion of the blade, t = 0.4 s

radius of the circle moved by the blade, r = 0.45 m

The angular speed of the blade in radian is calculated as;

`\omega = (\theta)/(t) * (\pi \ radian)/(180^0) \\\\\omega = (90.08 ^0)/(0.4 \ s) * (\pi \ radian)/(180^0) \\\\\omega = 3.93 \ rad/s`

The centripetal acceleration is calculated as;

a = ω²r

a = (3.93)² x 0.45

a = 6.95 m/s²