2-30 할리데이 10판 솔루션 일반물리학

$$ \begin{aligned} v_0 &= 146\ut{km/h} \\ &= 146\ut{km/h}\cdot\frac{1000\ut{m}}{1\ut{km}}\cdot\frac{1\ut{h}}{3600\ut{s}} \\ &= \frac{365}{9}\ut{m/s} \end{aligned}$$ $$ \begin{aligned}\\ v &= 90\ut{km/h} \\ &= 90\ut{km/h}\cdot\frac{1000\ut{m}}{1\ut{km}}\cdot\frac{1\ut{h}}{3600\ut{s}} \\ &= 25\ut{m/s} \end{aligned}$$ $$ \begin{cases} a &= -5.2\ut{m/s^2} \\ v_0 &= \frac{365}{9}\ut{m/s} \\ v &= 25\ut{m/s} \end{cases}$$
(a) $$t= ? $$ $$ v=v_0+at, $$ $$ \begin{aligned} t&=\frac{v-v_0}{a} \\ &=\frac{25\ut{m/s}-\frac{365}{9}\ut{m/s}}{-5.2\ut{m/s^2}} \\ &=\frac{350}{117}\ut{s} \\ &\approx 2.991452991452992\ut{s} \\ &\approx 3.0\ut{s} \end{aligned} $$
(b) $$ x(t) = ? $$ $$ S = v_0t+\frac{1}{2}at^2, $$ $$ \begin{aligned} x(t) &= S = (\frac{365}{9})t+\frac{1}{2}(-5.2)t^2 \\ &= \frac{365}{9}t-\frac{13}{5}t^2 \end{aligned}$$ $$\begin{aligned} v(t) &= \dot x = \dxt{x} \\ &= \dt(\frac{365}{9}t-\frac{13}{5}t^2) \\ &= \frac{365}{9}-\frac{26}{5}t \end{aligned} $$