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A shaft is turning at \({65}{r}{a}{d}/{s}\) at time zero. Thereafter, angular acceleration is given by \(\alpha=-{10}{r}{a}{d}/{s}^{{{2}}}-{5}{t}{r}{a}{d}/{s}^{{{2}}}\)
Where \({t}\) is the elapsed time
(a). Find its angular speed at \({t}={3.0}\) s
(b). How much angle does it turn in these \({3}{s}\)?


(a). \(\int{d}\omega=\int\alpha{\left.{d}{t}\right.}\)
(b). \({\int_{{{65}}}^{{\omega}}}{d}\omega={\int_{{{0}}}^{{{t}}}}{\left(-{10}-{5}{t}\right)}{\left.{d}{t}\right.}\)
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