11판/2. 직선운동

2-52 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 1. 18. 15:20
{t0:Startt1:Max Speedt2:End \begin{cases} t_0&:\text{Start}\\ t_1&:\text{Max Speed}\\ t_2&:\text{End}\\ \end{cases} {v1=11.00[m/s]S01=12.0[m]S02=100[m] \begin{cases} v_{1}&=11.00\ut{m/s}\\ S_{0\rarr1}&=12.0\ut{m}\\ S_{0\rarr2}&=100\ut{m} \end{cases} (a)\ab{a} S=12(v+v0)t, S=\frac{1}{2}(v+v_0)t, S01=12(v1+v0)t01=12(11+0)t01t01=2S0111 \begin{aligned} S_{0\rarr1}&=\frac{1}{2}(v_1+v_0)t_{0\rarr1}\\ &=\frac{1}{2}(11+0)t_{0\rarr1}\\ t_{0\rarr1}&=\frac{2S_{0\rarr1}}{11}\\ \end{aligned} S12=v1t12100S01=11t12t12=100S0111 \begin{aligned} S_{1\rarr2}&=v_1t_{1\rarr2}\\ 100-S_{0\rarr1}&=11t_{1\rarr2}\\ t_{1\rarr2}&=\frac{100-S_{0\rarr1}}{11}\\ \end{aligned} Σt=t01+t12=(2S0111+100S0111)=S01+10011=(12)+10011=11211[s]10.18181818181818[s]10.2[s] \begin{aligned} \Sigma t&=t_{0\rarr1}+t_{1\rarr2}\\ &=\(\frac{2S_{0\rarr1}}{11}+\frac{100-S_{0\rarr1}}{11}\)\\ &=\frac{S_{0\rarr1}+100}{11}\\ &=\frac{(12)+100}{11}\\ &=\frac{112}{11}\ut{s}\\ &\approx 10.18181818181818\ut{s}\\ &\approx 10.2\ut{s}\\ \end{aligned} (b)\ab{b} Σt=S01+100119.9=S01+10011S01=8910[m]=8.9[m] \begin{aligned} \Sigma t&=\frac{S_{0\rarr1}+100}{11}\\ 9.9&=\frac{S_{0\rarr1}+100}{11}\\ S_{0\rarr1}&=\frac{89}{10}\ut{m}\\ &=8.9\ut{m}\\ \end{aligned}