2-40 할리데이 10판 솔루션 일반물리학

$$\begin{aligned} v_0&=55\ut{km/h}\cdot\frac{1000\ut{m}}{1\ut{km}}\cdot\frac{1\ut{h}}{3600\ut{s}} \\ &=\frac{275}{18}\ut{m/s}\\ \end{aligned}$$ $$\begin{cases} v_0 = \frac{275}{18}\ut{m/s} \\ a=-5.18\ut{m/s^2} \\ T = 0.75\ut{s} \\ \end{cases}$$
(a) $$\begin{cases} x_{A3} = 40\ut{m} \\ t_A = 2.8\ut{s} \\ \end{cases}$$ $$\title{Case A Still Run}$$ $$\begin{aligned} \Delta x&=vt, \\ \Delta x_A&=v_0t_A \\ &=(\frac{275}{18}\ut{m/s})(2.8\ut{s}) \\ &=\frac{385}{9}\ut{m}\\ &\approx 42.77777777777778\ut{m} > 40\ut{m}\\ \end{aligned}$$ $$\therefore \title{Case A CAN Still Run}$$ $$\title{Case A Brake}$$ $$\begin{aligned} \Delta x_S &=v_0T \\ &= \frac{275}{18}\ut{m/s}\cdot0.75\ut{s} \\ &= \frac{275}{24}\ut{m} \end{aligned}$$ $$\begin{aligned} t_a &= t_A-0.75\ut{s} \\ &= 2.8\ut{s}-0.75\ut{s} \\ &= 2.05\ut{s} \\ \end{aligned}$$ $$\begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2, \\ \Delta x_a &= v_0t_a+\frac{1}{2}at_a^2 \\ &= (\frac{275}{18}\ut{m/s})(2.05\ut{s})+\frac{1}{2}(-5.18\ut{m/s^2})(2.05\ut{s})^2 \\ &= \frac{7356589}{360000}\ut{m}\\ \end{aligned}$$ $$\begin{aligned} x_A &= x_S + x_a \\ &= \frac{275}{24}\ut{m}+\frac{7356589}{360000}\ut{m} \\ &= \frac{11481589}{360000}\ut{m} \\ &\approx 31.89330277777778\ut{m} < 40\ut{m} \end{aligned}$$ $$\therefore \title{Case A CAN Brake}$$ $$\title{Case A Both Possible}$$
(b)
$$\begin{cases} x_{B3} = 32\ut{m} \\ t_B = 1.8\ut{s} \end{cases}$$ $$\title{Case B Still Run}$$ $$\begin{aligned} \Delta x&=vt, \\ \Delta x_B&=v_0t_B \\ &=(\frac{275}{18}\ut{m/s})(1.8\ut{s}) \\ &=\frac{55}{2}\ut{m}\\ &= 27.5\ut{m} \ngtr 32\ut{m}\\ \end{aligned}$$ $$\therefore \title{Case B can NOT Still Run}$$ $$\title{Case B Brake}$$ $$\begin{aligned} t_b &= t_B-0.75\ut{s} \\ &= 1.8\ut{s}-0.75\ut{s} \\ &= 1.05\ut{s} \\ \end{aligned}$$ $$\begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2, \\ \Delta x_b &= v_0t_b+\frac{1}{2}at_b^2 \\ &= (\frac{275}{18}\ut{m/s})(1.05\ut{s})+\frac{1}{2}(-5.18\ut{m/s^2})(1.05\ut{s})^2 \\ &= \frac{1582343}{120000}\ut{m}\\ \end{aligned}$$ $$\begin{aligned} x_B &= x_S + x_b \\ &= \frac{275}{24}\ut{m}+\frac{1582343}{120000}\ut{m} \\ &= \frac{985781}{40000}\ut{m} \\ &= 24.644525\ut{m} < 32\ut{m} \end{aligned}$$ $$\therefore \title{Case B CAN Brake}$$ $$\title{Case B Must Brake}$$