2-42 할리데이 10판 솔루션 일반물리학

$$\begin{cases} D = \text{Driver} \\ C = \text{Cop} \\ t_1 = \text{Driver Look Cop Speed Down}\\ t_2 = \text{Driver Brake Start}\\ t_3 = \text{Car Crash}\\ \end{cases}$$ $$\begin{aligned} v_0 &= v_{D0}= v_{C0}= 120\ut{km/h} \\ &= 120\ut{km/h}\cdot\frac{1000\ut{m}}{1\ut{km}}\cdot\frac{1\ut{h}}{3600\ut{s}} \\ &= \frac{100}{3}\ut{m/s} \\ \end{aligned}$$ $$\begin{cases} x_{D0} = 0 \\ x_{C0} = 25\ut{m} \\ v_0 = v_{D0}= v_{C0}= \frac{100}{3}\ut{m/s} \\ t_1 = 2\ut{s} \\ t_{1 \to 2} = 0.4\ut{s} \\ a=a_C = a_D = -5.0 \ut{m/s^2} \\ \end{cases}$$
(a) $x_{C2}-x_{D2}=?$ $$\begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2,\\ \Delta x_{C0\to2} &= v_0t_{0 \to 2}+\frac{1}{2}at_{0 \to 2}^2\\ x_{C2} &= x_{C0}+v_0t_{0 \to 2}+\frac{1}{2}at_{0 \to 2}^2\\ &= (25\ut{m})+(\frac{100}{3}\ut{m/s})(2.4\ut{s})+\frac{1}{2}(-5.0 \ut{m/s^2})(2.4\ut{s})^2\\ &=\frac{453}{5}\ut{m} \end{aligned}$$ $$\begin{aligned} \Delta x &= vt,\\ \Delta x_{D0\to2} &= v_0t_{0 \to 2}\\ x_{D2}&= (\frac{100}{3}\ut{m/s})(2.4\ut{s})\\ &= 80\ut{m} \end{aligned}$$ $$\begin{aligned} \Ans &= x_{C2}-x_{D2}\\ &= \frac{453}{5}\ut{m}-80\ut{m}\\ &= \frac{53}{5} \ut{m}\\ &= 10.6 \ut{m} \end{aligned}$$
(b) $v_{D3}=?$ $$\Delta x = v_0t+\frac{1}{2}at^2,$$ $$\begin{aligned} \Delta x_{D0\to3} &= \Delta x_{D0\to2}+\Delta x_{D2\to3}\\ x_{D3}&=\(v_0t_2\)+\(v_0t_{2\to3}+\frac{1}{2}at_{2\to3}^2\)\\ &=v_0t_{0\to3}+\frac{1}{2}at_{2\to3}^2\\ &=v_0t_3+\frac{1}{2}a(t_3-t_2)^2\\ \end{aligned}$$ $$\begin{aligned} \Delta x_{C0\to3}&=v_0t_{0\to3}+\frac{1}{2}at_{0\to3}^2\\ x_{C3}-x_{C0}&=v_0t_3+\frac{1}{2}at_3^2\\ x_{C3}&=x_{C0}+v_0t_3+\frac{1}{2}at_3^2\\ \end{aligned}$$ $$\begin{aligned} x_{D3}&=x_{C3},\\ v_0t_3+\frac{1}{2}a(t_3-t_2)^2&=x_{C0}+v_0t_3+\frac{1}{2}at_3^2\\ \frac{1}{2}a(\underset{x}{t_3}-t_2)^2&=x_{C0}+\frac{1}{2}a\underset{x}{t_3}^2\\ \end{aligned}$$ $$\text{(Underset x Means Unknown Value)}$$ $$\begin{aligned} t_3 &= \frac{t_2}{2}-\frac{x_{C0}}{at_2}\\ &=\frac{2.4\ut{s}}{2}-\frac{25\ut{m}}{(-5.0 \ut{m/s^2})(2.4\ut{s})}\\ &=\frac{197}{60}\ut{s}\\ \end{aligned}$$ $$\begin{aligned} v&=v_0+at,\\ v_{D3}&=v_0+at_3\\ &=\frac{100}{3}\ut{m/s}+\(-5.0 \ut{m/s^2}\)\(\frac{197}{60}\ut{s}\)\\ &=\frac{203}{12}\ut{m/s}\\ &\approx16.91666666666667\ut{m/s}\\ &\approx17\ut{m/s} \end{aligned}$$