Question

The blade of a windshield wiper moves through an angle of 90.0° in 0.397 s. The tip of the blade moves on the arc of a circle that has a radius of 0.376 m. What is the magnitude of the centripetal acceleration of the tip of the blade?

Answer 1

`a=5.88\ m/s^2`

Step-by-step explanation:

Given that,

The blade of a windshield wiper moves through an angle of 90.0° in 0.397 s

The radius of the circlular path is 0.376 m

We need to find the magnitude of the centripetal acceleration of the tip of the blade. Let it is a. The formula of the centripetal acceleration is given by :

`a=(v^2)/(r)`

v is velocity,
`v=(2\pi r)/(t\\)`

It moves through an angle of 90 degrees, it means
`v=((1)/(4)* 2\pi r)/(t\\)`

So,

`a=((((1)/(4)* 2\pi r)/(t\\))^2)/(r)\\\\=(\pi^2 r)/(4t^2)\\\\=(\pi^2 * 0.376)/(4(0.397 )^2)\\\\=5.88\ m/s^2`

So, the magnitude of the centripetal acceleration of the tip of the blade is
`5.88\ m/s^2`.