10판/2. 직선운동

2-21 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 7. 24. 04:57

v0=130[km/h]=130[km/h]1000[m]1[km]1[h]3600[s]=3259[m/s] \begin{aligned} v_0 &= 130\ut{km/h} \\&= 130\ut{km/h}\cdot\frac{1000\ut{m}}{1\ut{km}}\cdot\frac{1\ut{h}}{3600\ut{s}} \\ &= \frac{325}{9}\ut{m/s} \end{aligned} {v0=3259[m/s]v=0S=210[m]a=Constant \begin{cases} v_0 &= \frac{325}{9}\ut{m/s} \\ v &= 0 \\ S &= 210\ut{m} \\ a &= \Cons \end{cases}
(a) Ans=a=?\Ans = \abs{a} =? 2aS=v2v02,2aS = v^2-v_0^2, Ans=a=v2v022S=(0)2(3259[m/s])22(210[m])=211256804[m/s2]3.104791299235744[m/s2]3.10[m/s2] \begin{aligned} \Ans &= \abs{a} = \abs{\frac{v^2-v_0^2}{2S}} \\ &= \abs{\frac{(0)^2-\left(\frac{325}{9}\ut{m/s}\right)^2}{2(210\ut{m})}} \\ &= \frac{21125}{6804}\ut{m/s^2} \\ &\approx 3.104791299235744\ut{m/s^2} \\ &\approx 3.10\ut{m/s^2} \end{aligned}
(b) S=12(v+v0)t,S=\frac{1}{2}(v+v_0)t, t=2Sv+v0=2210[m]0+3259[m/s]=75665[s]11.63076923076923[s]11.6[s] \begin{aligned} t&= \frac{2S}{v+v_0} \\ &= \frac{2\cdot210\ut{m}}{0+\frac{325}{9}\ut{m/s}} \\ &= \frac{756}{65}\ut{s} \\ &\approx 11.63076923076923\ut{s} \\ &\approx 11.6\ut{s} \end{aligned}