10-38 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \theta(t\ut{s})&=4.0t-5.0t^2+t^3\ut{rad}\\ \end{cases} $$ $$\ab{a,b}$$ $$ \begin{aligned} \omega(t)&=\dyt{\theta}\\ &=\dt(4.0t-5.0t^2+t^3)\\ &=4 - 10 t + 3 t^2\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} \omega(2)&=-4.0\ut{rad/s} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \omega(5)&=29\ut{rad/s} \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \bar \alpha_{2\rarr5}&=\frac{\Sigma \Delta \omega}{\Sigma \Delta t}\\ &=11\ut{rad/s^2} \end{aligned} $$ $$\ab{d,e}$$ $$ \begin{aligned} \alpha(t)&=\dyt{\omega}\\ &=\dt(4 - 10 t + 3 t^2)\\ &=- 10 + 6 t\\ \end{aligned} $$ $$\ab{d}$$ $$ \alpha(2)=2.0\ut{rad/s^2} $$ $$\ab{d}$$ $$ \alpha(5)=2.0\times10\ut{rad/s^2} $$