2-3 할리데이 11판 솔루션 일반물리학

$$\begin{cases} t &= 0.50\ut{min}\\ v_{C0} &= 15\ut{km/h}\\ v_{C1} &= 45\ut{km/h}\\ v_{B0} &= 0\\ v_{B1} &= 30\ut{km/h}\\ \end{cases}$$ $$\begin{aligned} v&=v_0+at,\\ \end{aligned}$$ $$\ab{a}$$ $$\begin{aligned} a&=\frac{v-v_0}{t}\\ &=\frac{(45\ut{km/h})-(15\ut{km/h})}{0.50\ut{min}}\cdot\frac{60\ut{min}}{1\ut{h}} \\ &=3600\ut{km/h^2}\\ &=3600\ut{km/h^2}\cdot\frac{1000\ut{m}}{1\ut{km}}\cdot\(\frac{1\ut{h}}{3600\ut{sec}}\)^2\\ &=\frac{5}{18}\ut{m/s^2}\\ &\approx0.2777777777777778\ut{m/s^2}\\ &\approx0.28\ut{m/s^2}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} a&=\frac{v-v_0}{t}\\ &=\frac{(30\ut{km/h})-(0)}{0.50\ut{min}}\cdot\frac{60\ut{min}}{1\ut{h}}\\ &=3600\ut{km/h^2}\\ &=3600\ut{km/h^2}\cdot\frac{1000\ut{m}}{1\ut{km}}\cdot\(\frac{1\ut{h}}{3600\ut{sec}}\)^2\\ &=\frac{5}{18}\ut{m/s^2}\\ &\approx0.2777777777777778\ut{m/s^2}\\ &\approx0.28\ut{m/s^2}\\ \end{aligned}$$