2-65 할리데이 10판 솔루션 일반물리학

$$\begin{cases} A=\text{Head}\\ B=\text{Body}\\ \end{cases} $$
(a)$v_{A160}=?$ $$\begin{cases} v_0=0\\ a_A = \text{Linear} & \\ a_A(t) = 0 &(t<110)\\ a_A(160) = 90 & \\ \end{cases} $$ $$ \begin{aligned} (a_A-0)&=\frac{90-0}{160-110}(t-110)\\ a_A&=\frac{9}{5}(t-110)\\ \end{aligned} $$ $$ \begin{aligned} v(M)-v(N) &= \int^M_N a \dd t,\\ v(160)-v(110) &= \int^{160}_{110} a_A \dd t\\ v(160) &= \int^{160}_{110} \frac{9}{5}(t-110) \dd t\\ &=\[\frac{9}{5} \left(\frac{t^2}{2}-110t\right)\]^{160}_{110} \\ &=\[\frac{9}{5} \left(\frac{(160)^2}{2}-110(160)\right)\]-\[\frac{9}{5} \left(\frac{(110)^2}{2}-110(110)\right)\]\\ &=[-8640]-[-10890]\\ &=2250\ut{m/s} \end{aligned} $$
(b)$v_{B100\to120}=?$ $$\begin{cases} v_0=0\\ a_B = \text{Linear} & \\ a_B(t) = 0 &(t<40)\\ a_B(t) = 50 &(100<t<120)\\ \end{cases} $$ $$ \begin{aligned} (a_B-0)&=\frac{50-0}{100-40}(t-40)\\ a_B&=\frac{5 }{6}(t-40)\\ \end{aligned} $$ $$ \begin{aligned} v(M)-v(N) &= \int^M_N a \dd t,\\ v(100)-v(40) &= \int^{100}_{40} a_B \dd t\\ v(100) &= \int^{100}_{40} \frac{5 }{6}(t-40) \dd t\\ &=\[\frac{5}{6} \left(\frac{t^2}{2}-40t\right)\]^{100}_{40}\\ &=\[\frac{5}{6} \left(\frac{(100)^2}{2}-40(100)\right)\]-\[\frac{5}{6} \left(\frac{(40)^2}{2}-40(40)\right)\]\\ &=\[\frac{2500}{3}\]-\[-\frac{2000}{3}\]\\ &=1500\ut{m/s} \end{aligned} $$