11판/2. 직선운동

2-19 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 1. 11. 21:33
{d=20.0[m]vp=14.0[m/s]D23=200[m]D12=300[m] \begin{cases} d&=20.0\ut{m}\\ v_p&=14.0\ut{m/s}\\ D_{23}&=200\ut{m}\\ D_{12}&=300\ut{m} \end{cases} (a)\ab{a} Δt=dv=20.0[m]14.0[m/s]=107[s]1.428571428571429[s]1.43[s] \begin{aligned} \Delta t &= \frac{d}{v}\\ &=\frac{20.0\ut{m}}{14.0\ut{m/s}}\\ &=\frac{10}{7}\ut{s}\\ &\approx1.428571428571429\ut{s}\\ &\approx1.43\ut{s} \end{aligned} (b)\ab{b} {tr=0.500[s]a=5.00[m/s2]v0=0tAns=tr+ΔtS=D12d=280[m] \begin{cases} t_r&=0.500\ut{s}\\ a&=5.00\ut{m/s^2}\\ v_0&=0\\ t_\Ans &=t_r+\Delta t\\ S&=D_{12}-d=280\ut{m} \end{cases} S=v0t+12at2,280=(0)Δt+12(5.00)(Δt)2 \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ 280&=(0)\Delta t+\frac{1}{2}(5.00)(\Delta t)^2\\ \end{aligned} Δt=47[s]\Delta t =4\sqrt{7}\ut{s} tAns=0.500[s]+47[s]=(12+47)[s]11.08300524425836[s]11.1[s] \begin{aligned} t_\Ans &=0.500\ut{s}+4\sqrt{7}\ut{s}\\ &=\(\frac{1}{2}+4\sqrt{7}\)\ut{s}\\ &\approx 11.08300524425836\ut{s}\\ &\approx 11.1\ut{s}\\ \end{aligned}