4-64 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} A:(5.00,3.00)\ut{m}\\ B:(12.0,18.0)\ut{m}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} &\text{dirction of } \vec a \\ &=\text{dirction of } \vec v \\ &=\text{dirction of } \Delta\vec x \end{aligned} $$ $$ \begin{aligned} \Ans &= \frac{a_y}{a_x}\\ &= \frac{v_y}{v_x}\\ &= \frac{\Delta y}{\Delta x}\\ &= \frac{18-3}{12-5}\\ &= \frac{15}{7}\\ &\approx 2.142857142857143\\ &\approx 2.14\\ \end{aligned} $$ $$ S=v_0t+\frac{1}{2}at^2, $$ $$ \begin{cases} S_2&=v_0t_2+\frac{1}{2}a{t_2}^2\\ S_1&=v_0t_1+\frac{1}{2}a{t_1}^2\\ \end{cases} $$ $$ \begin{cases} S_2&=(0)(2t_1)+\frac{1}{2}a(2t_1)^2\\ S_1&=(0)t_1+\frac{1}{2}a{t_1}^2\\ \end{cases} $$ $$ \begin{cases} S_2&=\frac{1}{2}a(2t_1)^2\\ S_1&=\frac{1}{2}a{t_1}^2\\ \end{cases} $$ $$ \begin{aligned} \frac{S_2}{S_1}&=\frac{(2t_1)^2}{{t_1}^2}\\ &=4\\ \therefore S_2 &= 4S_1 \end{aligned} $$ $$ \begin{aligned} \Delta \vec r_1 &= 7\i+15\j\\ \Delta \vec r_2 &= 4\Delta \vec r_1\\ &=28\i+60\j\\ \vec r_2 &= \vec r_0 + \Delta \vec r_2\\ &=5\i+3\j+28\i+60\j\\ &=33\i+63\j \end{aligned} $$ $$\therefore (33.0,63.0)$$