2-24 할리데이 10판 솔루션 일반물리학

$$ \begin{aligned} \\ S &= 1.0\ut{mm}-5.0\ut{\mu m} \\ &= 1.0\ut{mm}\frac{1\ut{m}}{10^3\ut{mm}}-5.0\ut{\mu m}\frac{1\ut{m}}{10^6\ut{\mu m}} \\ &= \frac{199}{200000}\ut{m} \\ &= 9.95\times 10^{-4}\ut{m} \end{aligned} $$ $$ \begin{cases} v_0 &= 1.6\ut{m/s} \\ S &= 9.95\times 10^{-4}\ut{m} \\ v &= 0 \\ a &= Const. \\ g &= 9.80665 \ut{m/s^2} \end{cases} $$
(a) t=? $$S = \frac{1}{2}(v+v_0)t ,$$ $$ \begin{aligned} \\ t &= \frac{2S}{v+v_0} \\ &= \frac{2 \cdot 9.95\times 10^{-4}\ut{m}}{0+1.6\ut{m/s}} \\ &= \frac{199}{160000}\ut{s} \\ &= 0.00124375\ut{s} \\ &\approx 1.2\ut{ms} \end{aligned} $$
(b) $$ \frac{a}{g} = ? $$ $$ 2aS = v^2-v_0^2 ,$$ $$ \begin{aligned} \\ \frac{a}{g} &= \frac{v^2-v_0^2}{2gS} \\ &= \frac{0^2-(1.6\ut{m/s})^2}{2(9.80665 \ut{m/s^2})(9.95\times 10^{-4}\ut{m})} \\ &= -\frac{5120000000}{39030467} \\ &\approx -131.1795731268089 \\ &\approx -1.3 \times 10^{2} \end{aligned} $$ $$ \therefore a \approx -1.3 \times 10^{2}g$$