10판/2. 직선운동

2-15 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 7. 23. 20:15

x=18t+5.0t2x = 18t + 5.0t^2
(a) Ans=x˙(2)=? \Ans = \dot x(2) =? x˙= ⁣dx ⁣dt= ⁣d ⁣dt(18t+5.0t2)=18+10tx˙(2)=18+10(2)=38[m/s] \begin{aligned} \dot x &= \dxt{x} \\&= \dt(18t + 5.0t^2) \\&= 18 + 10 t \\ \dot x(2)&=18 + 10 (2) \\ &= 38\ut{m/s} \end{aligned}
(b) Ans=x(t2)x(t1)t2t1=?=x(3.0)x(2.0)3.02.0={18(3.0)+5.0(3.0)2}{18(2.0)+5.0(2.0)2}=43[m/s] \begin{aligned} \Ans &= \frac{x(t_2)-x(t_1)}{t_2-t_1} =? \\ &= \frac{x(3.0)-x(2.0)}{3.0-2.0} \\ &= \bra{18(3.0) + 5.0(3.0)^2}-\bra{18(2.0) + 5.0(2.0)^2} \\ &= 43\ut{m/s} \end{aligned}