4-56 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \vec r&=4.0t\i+4.0\sin\cfrac{\pi t}{4.0}\j \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec r(t)&=4t\i+4\sin\frac{\pi t}{4}\j,\\ \end{aligned} $$ $$ \begin{aligned} \vec r(1.0)&=4\i+2\sqrt2\j\ut{m}\approx 4.0\i+ 2.8\j\ut{m}\\ \vec r(2.0)&=8\i+4\j\ut{m}=8.0\i+4.0\j\ut{m}\\ \vec r(3.0)&=12\i+2\sqrt2\j\ut{m}\approx 12\i+ 2.8\j\ut{m}\\ \vec r(4.0)&=16\i+0\j\ut{m} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \vec v(t)&=\dxt{\vec r(t)}\\ &=\dt\(4t\i+4\sin\frac{\pi t}{4}\j\)\\ &=4\i+\pi\cos\frac{\pi t}{4}\j,\\ \end{aligned} $$ $$ \begin{aligned} \vec v(1.0)&=4\i+\frac{\pi}{\sqrt2}\j\ut{m/s}\approx 4.0\i+ 2.2\j\ut{m/s}\\ \vec v(2.0)&=4\i+0\j\ut{m/s}=4.0\i+0\j\ut{m/s}\\ \vec v(3.0)&=4\i-\frac{\pi}{\sqrt2}\j\ut{m/s}\approx 4.0\i- 2.2\j\ut{m/s}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \vec a(t)&=\dxt{\vec v(t)}\\ &=\dt\(4\i+\pi\cos\frac{\pi t}{4}\j\)\\ &=0\i-\frac{\pi ^2}{4} \sin \frac{\pi t}{4}\j,\\ \end{aligned} $$ $$ \begin{aligned} \vec a(1.0)&=0\i-\frac{\pi^2}{4\sqrt2}\j\ut{m/s^2}\approx0\i-1.7\j\ut{m/s^2}\\ \vec a(2.0)&=0\i-\frac{\pi^2}{4}\j\ut{m/s^2}\approx0\i-2.5\j\ut{m/s^2}\\ \vec a(3.0)&=0\i-\frac{\pi^2}{4\sqrt2}\j\ut{m/s^2}\approx0\i-1.7\j\ut{m/s^2}\\ \end{aligned} $$