2-44 할리데이 11판 솔루션 일반물리학

$$y=(4.0\ut{cm})\sin\frac{\pi t}{4}$$ $$\ab{a}$$ $$\begin{aligned} \bar v_{0\rarr2}&=\frac{\Sigma \Delta y_{0\rarr2}}{\Sigma t_{0\rarr2}}\\ &=\frac{y(2)-y(0)}{2}\\ &=\frac{\(4\sin\dfrac{\pi\cdot2}{4}\)-\(4\sin\dfrac{\pi\cdot0 }{4}\)}{2}\\ &=\frac{\(4\)-\(0\)}{2}\ut{cm/s}\\ &=2\ut{cm/s}\\ &= 2.0\ut{cm/s} \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} v&=\dxt{y}\\ &=\dt\(4\sin\frac{\pi t}{4}\)\\ &=\pi\cos\frac{\pi t}{4}\\ \end{aligned}$$ $$\begin{aligned} v(0)&=\pi\cos\(\frac{\pi\cdot0}{4}\)\\ &=\pi\\ &\approx 3.141592653589793\ut{cm/s}\\ &\approx 3.1\ut{cm/s}\\ v(1)&=\pi\cos\(\frac{\pi\cdot1}{4}\)\\ &=\frac{\pi}{\sqrt 2}\\ &\approx 2.221441469079183\ut{cm/s}\\ &\approx 2.2\ut{cm/s}\\ v(2)&=\pi\cos\(\frac{\pi\cdot2}{4}\)\\ &=0\\ \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} \bar a_{0\rarr2} &= \frac{\Sigma \Delta v}{\Sigma \Delta t}\\ &= \frac{v(2)-v(0)}{2}\\ &= \frac{(0)-(\pi)}{2}\ut{cm/s^2}\\ &= -\frac{\pi}{2}\ut{cm/s^2}\\ &\approx -1.570796326794897\ut{cm/s^2}\\ &\approx -1.6\ut{cm/s^2}\\ \end{aligned}$$ $$\ab{d}$$ $$\begin{aligned} a&=\dxt{v}\\ &=\dt\(\pi\cos\frac{\pi t}{4}\)\\ &=-\frac{\pi ^2}{4} \sin \frac{\pi t}{4}\\ \end{aligned}$$ $$\begin{aligned} a(0)&=-\frac{\pi^2}{4} \sin\(\frac{\pi \cdot 0}{4}\)\\ &=0\\ a(1)&=-\frac{\pi^2}{4} \sin\(\frac{\pi \cdot 1}{4}\)\\ &=-\frac{\pi ^2}{4 \sqrt{2}}\ut{cm/s^2}\\ &\approx -1.74471604990972\ut{cm/s^2}\\ &\approx -1.7\ut{cm/s^2}\\ a(2)&=-\frac{\pi^2}{4} \sin\(\frac{\pi \cdot 2}{4}\)\\ &=-\frac{\pi ^2}{4}\ut{cm/s^2}\\ &\approx -2.46740110027234\ut{cm/s^2}\\ &\approx -2.5\ut{cm/s^2}\\ \end{aligned}$$