1-13 할리데이 10판 솔루션 일반물리학

$$ \text{put} \begin{cases} t_A(t_B)&=a_A\cdot t_B +b_A \\ t_B&=ref \\ t_C(t_B)&=a_C\cdot t_B +b_C \end{cases} $$ $$ \text{picture 1-4}, \begin{cases} t_A(125)&=312&=125a_A +b_A \\ t_A(290)&=512&=290a_A +b_A \\ t_C(25.0)&=92.0&=25.0a_C +b_C \\ t_C(200)&=142&=200a_C +b_C \end{cases} $$ $$ \therefore \begin{cases} t_A(t_B)&=\frac{40}{33}t_B +\frac{5818}{33} \\ t_B&=ref \\ t_C(t_B)&=\frac{2}{7}t_B +\frac{594}{7} \end{cases} $$
(a),(b) $\Delta t_A=600,$
(a) $\Delta t_A=600, \Delta t_B=?$ $$ \begin{aligned} \Delta t_A &= t_A(t_{B2})-t_A(t_{B1}) \\ &=\(\frac{40}{33} t_{B2} + \frac{5818}{33}\) - \(\frac{40}{33} t_{B1} + \frac{5818}{33}\) \\ &= \frac{40}{33}t_{B2}-\frac{40}{33}t_{B1} \\ &= \frac{40}{33}(t_{B2}-t_{B1}) \\ &= \frac{40}{33}(\Delta t_B) \end{aligned} $$ $$ \begin{aligned} \\ \therefore \Delta t_B&=\frac{33}{40}\Delta t_A \\ &=\frac{33}{40}\times600 \\ &=495\ut{s} \end{aligned} $$
(b) $\Delta t_A=600, \Delta t_C=?$
$$ \begin{aligned} \Delta t_C &= t_C(t_{B2})-t_C(t_{B1}) \\ &=\(\frac{2}{7}t_{B2} +\frac{594}{7}\) - \(\frac{2}{7}t_{B1} +\frac{594}{7} \) \\ &= \frac{2}{7}t_{B2}-\frac{2}{7}t_{B1} \\ &= \frac{2}{7}(t_{B2}-t_{B1}) \\ &= \frac{2}{7}(\Delta t_B) \\ &= \frac{2}{7}\times 495 \\ &= \frac{990}{7}\ut{s} \\ &\approx 141.42857142857142857\ut{s} \\ &\approx 141\ut{s} \end{aligned} $$
(c)$t_A=400, t_B=?$
$$ \begin{aligned} t_A(t_B) &=\frac{40}{33}t_B +\frac{5818}{33} = 400\ut{s} \end{aligned} $$ $$ \begin{aligned} \\ t_B &= \frac{3691}{20}\ut{s} \\ &= 184.55\ut{s} \\ &\approx 185\ut{s} \end{aligned} $$
(d)$t_C=15.0, t_B=?$
$$ \begin{aligned} t_C(t_B)&=\frac{2}{7}t_B +\frac{594}{7} = 15.0\ut{s} \end{aligned} $$ $$ \begin{aligned} \\ t_B &= -\frac{489}{2}\ut{s} \\ &= -244.5\ut{s} \\ &\approx -245\ut{s} \end{aligned} $$