A 0.30-m radius automobile tire accelerates from rest at a constant 3 rad/s².What is the centripetal acceleration of a point on the outer edge of the tire after 5 s? A) 5.7 m/s² B) 67.5 m/s² C) 6.8 m/s² D) 4.5 m/s² E) 45.0 m/s²

Question

Quizplus · Accepted Answer

Centripetal acceleration is given by:$$a_c = rω^2$$where:r is the radius of the circleω is the angular velocityIn this case, the radius is 0.30 m and the angular acceleration is 3 rad/s^2. After 5 s, the angular velocity is:$$ω = αt = 3 rad/s^2 * 5 s = 15 rad/s$$So, the centripetal acceleration is:$$a_c = rω^2 = 0.30 m * (15 rad/s)^2 = 67.5 m/s^2$$

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