10-30 할리데이 11판 솔루션 일반물리학

$$ \put \begin{cases} a_t&:\text{Tangential Acceleration}\\ a_r&:\text{Rotational Acceleration}\\ \t &:\text{Unit Tangent Vector}\\ \r &:\text{Unit Radius Vector}\\ \end{cases} $$ $$ \begin{cases} v_i&=0\\ r&=26.0\ut{m}\\ a_t&=0.500\ut{m/s^2}\\ \Delta t&=8.00\ut{s} \end{cases} $$ $$v=v_0+at,$$ $$ \begin{aligned} v&=\frac{t}{2} \end{aligned} $$ $$ \begin{aligned} \vec a&=a_t\t+a_r\r\\ &=a_t\t+\frac{v^2}{r}\r\\ &=\frac{1}{2}\t+\frac{t^2}{104}\r\\ \end{aligned} $$ $$ \begin{aligned} \vec a(8)&=\frac{1}{2}\t+\frac{8}{13}\r\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} a(8)&=\sqrt{\(\frac{1}{2}\)^2+\(\frac{8}{13}\)^2}\\ &=\frac{5 \sqrt{17}}{26}\ut{m/s^2}\\ &\approx 0.7929049280033963\ut{m/s^2}\\ &\approx 0.793\ut{m/s^2}\\ &\approx 79.3\ut{cm/s^2}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \theta_{\vec a}&=\tan^{-1}\frac{a_r}{a_t}\\ &=\tan^{-1}\frac{16}{13}\\ &\approx 0.8884797719201486\ut{rad}\\ &\approx 0.888\ut{rad}\\ \end{aligned} $$