68. We note that after three seconds have elapsed (t2– t1= 3.00 s) the velocity (for this object in circular motion of period T) is reversed; we infer that it takes three seconds to reach the opposite side of the circle. Thus, T= 2(3.00 s) = 6.00 s.(a) Using Eq. 4-35, r= vT/2π, where 22(3.00 m/s)(4.00 m/s)5.00 m/sv=+=, we obtain 4.77 mr=. The magnitude of the object’s centripetal acceleration is therefore
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.