11판/2. 직선운동

2-27 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 1. 12. 18:44
{t=1.8[s]v1=1400[km/h]=35009[m/s]g=9.80665[m/s2] \begin{cases} t&=1.8\ut{s}\\ v_1&=1400\ut{km/h}=\cfrac{3500}{9}\ut{m/s}\\ g&=9.80665\ut{m/s^2} \end{cases} (a)\ab{a} v=v0+at,(35009)=(0)+a(1.8) \begin{aligned} v&=v_0+at,\\ \(\frac{3500}{9}\)&=(0)+a(1.8)\\ \end{aligned} a=1750081[m/s2]=500000002269539g22.03090583594289g22g \begin{aligned} a&=\frac{17500}{81}\ut{m/s^2}\\ &=\frac{50000000}{2269539}g\\ &\approx22.03090583594289g\\ &\approx22g\\ \end{aligned} (b)\ab{b} S=12(v+v0)t,=12{(35009)+0}(1.8)=350[m]=3.5×102[m] \begin{aligned} S&=\frac{1}{2}(v+v_0)t,\\ &=\frac{1}{2}\bra{\(\frac{3500}{9}\)+0}(1.8)\\ &=350\ut{m}\\ &=3.5\times10^2\ut{m} \end{aligned}