10판/2. 직선운동

2-24 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 7. 24. 21:03

S=1.0[mm]5.0[μm]=1.0[mm]1[m]103[mm]5.0[μm]1[m]106[μm]=199200000[m]=9.95×104[m] \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} {v0=1.6[m/s]S=9.95×104[m]v=0a=Const.g=9.80665[m/s2] \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=12(v+v0)t,S = \frac{1}{2}(v+v_0)t , t=2Sv+v0=29.95×104[m]0+1.6[m/s]=199160000[s]=0.00124375[s]1.2[ms] \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) ag=? \frac{a}{g} = ? 2aS=v2v02, 2aS = v^2-v_0^2 , ag=v2v022gS=02(1.6[m/s])22(9.80665[m/s2])(9.95×104[m])=512000000039030467131.17957312680891.3×102 \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} a1.3×102g \therefore a \approx -1.3 \times 10^{2}g