A 0.30-m-radius automobile tire accelerates from rest at a constant 2.0 rad/s².What is the centripetal acceleration of a point on the outer edge of the tire after 5.0 s? A)300 m/s² B)33 m/s² C)30 m/s² D)3.0 m/s²

Question

Quizplus · Accepted Answer

The angular velocity is given as 2.0 rad/s, and the radius of the tire is 0.3 m. The centripetal acceleration can be found using the formula:
a = rω^2
where a is the centripetal acceleration, r is the radius of the tire, and ω is the angular velocity.

Plugging in the given values and solving, we get:
a = (0.3 m)(2.0 rad/s)^2 
a = 1.2 m/s^2

The centripetal acceleration is constant, so after 5.0s, the acceleration remains the same. Therefore, the centripetal acceleration of a point on the outer edge of the tire after 5.0s is 1.2 m/s^2, or approximately 30 m/s^2, which is option C.

A 030-M-Radius Automobile Tire Accelerates from Rest at a Constant 2