11판/2. 직선운동

2-4 할리데이 11판 솔루션 일반물리학

짱세디럭스 2022. 10. 10. 20:39
문제의 유효숫자가 애매하여 다양하게 표현하였습니다. {v1=30[m/s]S1>2=160[m]v2=60[m/s] \begin{cases} v_1&=30\ut{m/s}\\ S_{1->2}&=160\ut{m}\\ v_2&=60\ut{m/s} \end{cases} (a)\ab{a} 2aS=v2v02, \begin{aligned} 2aS&=v^2-{v_0}^2,\\ \end{aligned} a=v2v022S=(60[m/s])2(30[m/s])22(160[m])=13516[m/s2]=8.4375[m/s2]8.4[m/s2]8[m/s2] \begin{aligned} a&=\frac{v^2-{v_0}^2}{2S}\\ &=\frac{(60\ut{m/s})^2-{(30\ut{m/s})}^2}{2(160\ut{m})}\\ &=\frac{135}{16}\ut{m/s^2}\\ &=8.4375\ut{m/s^2}\\ &\approx8.4\ut{m/s^2}\\ &\approx8\ut{m/s^2}\\ \end{aligned} (b)\ab{b} S=12(v+v0)t, \begin{aligned} S&=\frac{1}{2}(v+v_0)t,\\ \end{aligned} t=2Sv+v0=2(160[m])60[m/s]+30[m/s]=329[s]3.5555555555555554[s]3.6[s]4[s] \begin{aligned} t&=\frac{2S}{v+v_0}\\ &=\frac{2(160\ut{m})}{60\ut{m/s}+30\ut{m/s}}\\ &=\frac{32}{9}\ut{s}\\ &\approx3.5555555555555554\ut{s}\\ &\approx3.6\ut{s}\\ &\approx4\ut{s}\\ \end{aligned} (c)\ab{c} v=v0+at, \begin{aligned} v&=v_0+at,\\ \end{aligned} t=vv0a=60[m/s]30[m/s]13516[m/s2]=329[s]3.5555555555555554[s]3.6[s]4[s] \begin{aligned} t&=\frac{v-v_0}{a}\\ &=\frac{60\ut{m/s}-30\ut{m/s}}{\frac{135}{16}\ut{m/s^2}}\\ &=\frac{32}{9}\ut{s}\\ &\approx3.5555555555555554\ut{s}\\ &\approx3.6\ut{s}\\ &\approx4\ut{s}\\ \end{aligned} (d)\ab{d} 2aS=v2v02, \begin{aligned} 2aS&=v^2-{v_0}^2,\\ \end{aligned} S=v2v022a=(60[m/s])2(30[m/s])22(13516[m/s2])=160[m]=1.6×102[m]2×102[m] \begin{aligned} S&=\frac{v^2-{v_0}^2}{2a}\\ &=\frac{(60\ut{m/s})^2-{(30\ut{m/s})}^2}{2\(\cfrac{135}{16}\ut{m/s^2}\)}\\ &=160\ut{m}\\ &=1.6\times10^2\ut{m}\\ &\approx2\times10^2\ut{m}\\ \end{aligned}