A 0.40-m radius automobile tire accelerates from rest at a constant 2.0 rad/s² over a 5.0 s interval.What is the tangential component of acceleration for a point on the outer edge of the tire during the 5.0-s interval? A) 33 m/s² B) 0.80 m/s² C) 0.60 m/s² D) 0.30 m/s²

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Quizplus · Accepted Answer

The tangential acceleration is given by the formula $ a_t = \alpha \times r $, where $ \alpha = 2.0 \, \text{rad/s}^2 $ and $ r = 0.40 \, \text{m} $. Thus, $ a_t = 2.0 \times 0.40 = 0.80 \, \text{m/s}^2 $.

A 040-M Radius Automobile Tire Accelerates from Rest at a Constant