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Sagot :
Since we assume the accelerations are constant, the instantaneous acceleration is the same as the average acceleration, so in the first 6.80 seconds we have
[tex]\alpha_1 = \alpha_{\rm ave} = \dfrac{\Delta \omega}{\Delta t} = \dfrac{8.70\frac{\rm rev}{\rm s}}{6.80\,\rm s} \approx 1.28 \dfrac{\rm rev}{\mathrm s^2}[/tex]
In this time, the tub undergoes an angular displacement of
[tex]\theta_1 = \dfrac12 \alpha_1 (6.80\,\mathrm s)^2 \approx 29.6\,\mathrm{rev}[/tex]
In the next 19.0 seconds the tub has acceleration
[tex]\alpha_2 = \dfrac{-8.70\frac{\rm rev}{\rm s}}{19.0\,\rm s} \approx -0.458 \dfrac{\rm rev}{\mathrm s^2}[/tex]
and in this time, the tube undergoes an additional displacement of
[tex]\theta_2 = \left(8.70\dfrac{\rm m}{\rm s}\right)(19.0\,\mathrm s) + \dfrac12 \alpha_2 (19.0\,\mathrm s)^2 \approx 82.7\,\mathrm{rev}[/tex]
So the tub completes [tex]\theta_1+\theta_2 \approx \boxed{112\,\mathrm{rev}}[/tex].
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