4-19 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} v&=216\ut{km/h}=60.0\ut{m/s}\\ a&=0.050g\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} a_R&=\frac{v^2}{R}\\ R&=\frac{v^2}{a_R}\\ &=\frac{(60)^2}{0.050g}\\ &=\frac{72000}{g}\\ &=\frac{1440000000}{196133}\ut{m}\\ &\approx 7341.956733441083\ut{m}\\ &\approx 7.3\ut{km}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} a_R&=\frac{v^2}{R}\\ v&=\sqrt{a_R \cdot R}\\ &=\sqrt{0.050g \cdot 2300}\\ &=\sqrt{115g}\\ &=\frac{1}{20}\sqrt{\frac{4511059}{10}}\ut{m/s}\\ &\approx 33.58220883146313\ut{m/s}\\ &\approx 34\ut{m/s}\\ \end{aligned} $$