11판/2. 직선운동

2-59 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 1. 18. 18:42
{t=4.8[s]v0=0v1=60[km/h]=503[m/s] \begin{cases} t&=4.8\ut{s}\\ v_0&=0\\ v_1&=60\ut{km/h}=\frac{50}{3}\ut{m/s}\\ \end{cases} (a)\ab{a} v=v0+at,503=0+a(4.8) \begin{aligned} v&=v_0+at,\\ \frac{50}{3}&=0+a(4.8)\\ \end{aligned} a=12536[m/s2]3.472222222222222[m/s2]3.5[m/s2] \begin{aligned} a&=\frac{125}{36}\ut{m/s^2}\\ &\approx 3.472222222222222\ut{m/s^2}\\ &\approx 3.5\ut{m/s^2}\\ \end{aligned} (b)\ab{b} S=12(v+v0)t,=12(503+0)(4.8)=40[m] \begin{aligned} S&=\frac{1}{2}(v+v_0)t,\\ &=\frac{1}{2}\(\frac{50}{3}+0\)\cdot(4.8)\\ &=40\ut{m}\\ \end{aligned} (c)\ab{c} S=v0t+12at2,250=(0)t+12(12536)t2 \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ 250&=(0)t+\frac{1}{2}\(\frac{125}{36}\)t^2\\ \end{aligned} t=12[s] \begin{aligned} t&=12\ut{s} \end{aligned}