10판/2. 직선운동

2-35 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 7. 25. 07:44

{S=3[cm]=0.03[m]v0=4.00×105[m/s]v=6.00×107[m/s] \begin{cases} S &= 3\ut{cm}=0.03\ut{m} \\ v_0 &= 4.00 \times 10^5 \ut{m/s} \\ v &= 6.00 \times 10^7\ut{m/s} \end{cases} t=? t =? S=12(v+v0)t, S = \frac{1}{2}(v+v_0)t, t=2Sv+v0=2(0.03[m])6.00×107[m/s]+4.00×105[m/s]=33020000000[s]9.933774834437087×1010[s]9.93×1010[s]=993[ps] \begin{aligned} t &= \frac{2S}{v+v_0} \\ &= \frac{2(0.03\ut{m})}{6.00 \times 10^7\ut{m/s}+4.00 \times 10^5 \ut{m/s}} \\ &=\frac{3}{3020000000}\ut{s} \\ &\approx 9.933774834437087\times10^{-10}\ut{s} \\ &\approx 9.93\times10^{-10}\ut{s} = 993\ut{ps} \end{aligned}