2-29 할리데이 10판 솔루션 일반물리학

$$\begin{aligned} \max v &= 305\ut{m/min} \\ &= 305\ut{m/min}\frac{1\ut{min}}{60\ut{s}} \\ &= \frac{61}{12}\ut{m/s} \end{aligned}$$ $$\begin{cases} \max v &= v_M=\frac{61}{12}\ut{m/s} \\ \max S &= S_M=190 \ut{m} \\ |a| &= 1.22\ut{m/s^2} \\ v_0 &= 0 \end{cases}$$
(a) $$S_a =?$$ $$2aS = v^2-v_0^2,$$ $$\begin{aligned} S_a &= \frac{v_M^2}{2a}(\because v_0=0) \\ &= \frac{(\frac{61}{12}\ut{m/s})^2}{2(1.22\ut{m/s^2})} \\ &= \frac{1525}{144}\ut{m} \\ &\approx 10.59027777777778\ut{m} \\ &\approx 10.6\ut{m} \end{aligned}$$
(b) $$t=?$$ $$\begin{aligned} t &= t_{v+}+t_{v_M}+t_{v-} \\ &= t_{v_0 \to v_M}+t_{v_M}+t_{v_M \to v_0}(\because S_a<\frac{S_M}{2}) \\ &= 2t_{v_0 \to v_M}+t_{S_M - 2S_a} \ (t_{v_0 \to v_M}\overset{\text{def}}{=}t_b, S_M-2S_a \overset{\text{def}}{=} S_b ) \\ &= 2t_b+t_{S_b} \end{aligned}$$ $$v_M=v_0+at = at (\because v_0=0)$$ $$t_b = \frac{v_M}{a}$$ $$S_b=vt$$ $$t_{S_b} = \frac{S_b}{v_M}$$ $$\begin{aligned} \\ \therefore t&= 2t_b+t_{S_b} \\ &= 2\frac{v_M}{a}+\frac{S_b}{v_M} \\ &= 2\frac{v_M^2+aS_b}{av_M} \\ &= 2\frac{v_M^2+a(S_M-2S_a)}{av_M} \\ &= 2\frac{v_M^2+aS_M-2aS_a}{av_M} \\ &= 2\frac{v_M^2+aS_M-2a\frac{v_M^2}{2a}}{av_M} \\ &= 2\frac{v_M^2+aS_M-v_M^2}{av_M} \\ &= 2\frac{aS_M}{av_M} \\ &= 2\frac{S_M}{v_M} \\ &= 2\frac{190 \ut{m}}{\frac{61}{12}\ut{m/s}} \\ &= \frac{4560}{61}\ut{s} \\ &\approx 74.75409836065573\ut{s} \\ &\approx 74.8\ut{s} \\ \end{aligned}$$