4-18 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec a&=5.00\i+7.00\j\ut{m/s^2}\\ \vec v_0&=4.00\i\ut{m/s}\\ \end{cases}$$ $$t_{\Delta x=10}=?$$ $$\begin{aligned} \Delta r_x &= v_{x0}t+\frac{1}{2}a_xt^2\\ 10 &= 4t+\frac{1}{2}\cdot5t^2\\ t_{\Delta x=10}&=\frac{2}{5} \left(\sqrt{29}-2\right) \end{aligned}$$ $$v_{\Delta x=10}=?$$ $$\begin{aligned} \vec v&=\vec v_0 + \vec a t,\\ \vec v_{\Delta x=10}&=\vec v_0 + \vec a t_{\Delta x=10}\\ &=(4\i) + (5\i+7\j) \cdot \frac{2}{5} \left(\sqrt{29}-2\right)\\ &=2 \sqrt{29}\i+\frac{14}{5} \left(\sqrt{29}-2\right)\j\ut{m/s} \end{aligned}$$
(a) $\abs{\vec v_{\Delta x=10}}=?$ $$\begin{aligned} \abs{\vec v_{\Delta x=10}}&=\sqrt{\(2 \sqrt{29}\)^2+\left\{\frac{14}{5} \left(\sqrt{29}-2\right)\right\}^2}\ut{m/s}\\ &=\frac{2}{5} \sqrt{2342-196 \sqrt{29}}\ut{m/s}\\ &\approx14.34716807067729\ut{m/s}\\ &\approx14.3\ut{m/s}\\ \end{aligned}$$
(b) $\theta_{\vec v_{\Delta x=10}}=?$ $$\begin{aligned} \theta_{\vec v_{\Delta x=10}}&=\tan^{-1}\(\frac{\frac{14}{5} \left(\sqrt{29}-2\right)}{2 \sqrt{29}}\)\\ &=\tan ^{-1}\left(\frac{7}{5}-\frac{14}{5 \sqrt{29}}\right)\\ &\approx 0.7216847492576719\ut{rad}\\ &\approx 0.722\ut{rad}\\ \end{aligned}$$