10-19 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} v_x&=480\ut{km/h}=\frac{400}{3}\ut{m/s}\\ r&=1.5\ut{m}\\ \omega &=2000\ut{rev/min}=\frac{200\pi}{3}\ut{rad/s} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} v_t&=r\omega\\ &=100\pi\ut{m/s}\\ &\approx 3.141592653589793\times10^2\ut{m/s}\\ &\approx 3.1\times10^2\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} v&=\sqrt{{v_x}^2+{v_t}^2}\\ &\approx 3.4128261278399657\times10^2\ut{m/s}\\ &\approx 3.4\times10^2\ut{m/s}\\ \end{aligned} $$