11판/2. 직선운동

2-7 할리데이 11판 솔루션 일반물리학

짱세디럭스 2023. 6. 23. 22:25
(a)\ab{a} x5=x0+Δx05=0+05v ⁣dt=12(2[s]+4[s])8[m/s]+12(4[m/s]+8[m/s])1[s]=30[m] \begin{aligned} x_5&=x_0+\Delta x_{0\rarr5} \\ &=0+\int_0^5 v\dd t \\ &=\frac{1}{2}(2\ut{s}+4\ut{s})8\ut{m/s}+\frac{1}{2}(4\ut{m/s}+8\ut{m/s})1\ut{s}\\ &=30\ut{m} \end{aligned} (b)\ab{b} v5=4[m/s] v_5=4\ut{m/s} (c)\ab{c} a5= ⁣d ⁣dtv=8[m/s]2[s]=4[m/s2] \begin{aligned} a_5&=\dt v\\ &=\frac{-8\ut{m/s}}{2\ut{s}}\\ &=-4\ut{m/s^2} \end{aligned} (d)\ab{d} vˉ15=Δx1551=Δx05Δx014=3012144=7[m/s] \begin{aligned} \bar{v}_{1\rarr5}&=\frac{\Delta x_{1\rarr5}}{5-1}\\ &=\frac{\Delta x_{0\rarr5}-\Delta x_{0\rarr1}}{4}\\ &=\frac{30-\frac{1}{2}\cdot1\cdot4}{4}\\ &=7\ut{m/s} \end{aligned} (e)\ab{e} aˉ15=v5v151=4[m/s]4[m/s]4=0 \begin{aligned} \bar{a}_{1\rarr5}&=\frac{v_5-v_1}{5-1}\\ &=\frac{4\ut{m/s}-4\ut{m/s}}{4}\\ &=0 \end{aligned}