솔루션피아

$$\begin{cases} \vec a_1 = 3.66\hat j\ut{m}\\ \vec a_2:(1.83\ut{m},-\frac{\pi}{4})\\ \vec a_3:(0.91\ut{m},-\frac{3}{4}\pi)\\ \end{cases}$$ $$\begin{aligned} \vec a_2 &= 1.83\cos\(-\frac{\pi}{4}\)\hat i+1.83\sin\(-\frac{\pi}{4}\)\hat j\\ &= \frac{1.83}{\sqrt{2}}\hat i-\frac{1.83}{\sqrt{2}}\hat j\\ \end{aligned}$$ $$ \begin{aligned} \vec a_3 &= 0.91\cos\(-\frac{3}{4}\pi\)\hat i+0.91\sin\(-\frac..

(a) 20.0° $$20.0^\circ =20.0^\circ \cdot \frac{\pi\ut{rad}}{180^\circ}=\frac{\pi}{9}\ut{rad}\approx0.349\ut{rad}$$ (b) 50.0° $$50.0^\circ =50.0^\circ \cdot \frac{\pi\ut{rad}}{180^\circ}=\frac{5}{18}\pi\ut{rad}\approx0.873\ut{rad}$$ (c) 100° $$100^\circ =100^\circ \cdot \frac{\pi\ut{rad}}{180^\circ}=\frac{5}{9}\pi\ut{rad}\approx1.75\ut{rad}$$ (d) 0.330 rad $$ 0.330\ut{rad} = 0.330\ut{rad} \cdo..

$$\vec x = 15\ut{m}\hat i + 15\ut{m}\hat j + 10\ut{m}\hat k$$ $$x=?$$ $$\begin{aligned} x&=\abs{\vec x}\\ &=\sqrt{15^2+15^2+10^2}\ut{m}\\ &=5\sqrt{22}\ut{m}\\ &\approx 23.45207879911715\ut{m}\\ &\approx 23\ut{m}\\ \end{aligned}$$

$$\begin{cases} \vec r:(12\ut{m},30^\circ)\\ \vec r =r_x \hat i+ r_y \hat j\\ \end{cases}$$ (a)$r_x=?$ $$\begin{aligned} r_x&=r\cos\theta\\ &=12\cos30^\circ\ut{m}\\ &=6\sqrt{3}\ut{m} \end{aligned}$$ (a)$r_y=?$ $$\begin{aligned} r_x&=r\sin\theta\\ &=12\sin30^\circ\ut{m}\\ &=6\ut{m} \end{aligned}$$

$$\begin{cases} \vec a =a_x \hat i+ a_y \hat j\\ \vec a:(a,\theta)\\ a_x=\frac{1}{2}a\\ \end{cases}$$ $$\tan\theta = ?$$ $$a_x=a\cos\theta=\frac{1}{2}a$$ $$\cos\theta=\frac{1}{2}$$ $$\theta = \frac{\pi}{3}$$ $$\therefore \tan\theta = \tan\frac{\pi}{3}=\sqrt{3}$$

$$\begin{cases} \text{Particle 1}=A\\ \text{Particle 2}=B\\ \end{cases}$$ $$\begin{cases} x_A=6 .00t^2 + 3 .00t + 2.00\\ a_B=-8.00t\\ v_B(0)=15\ut{m/s}\\ \end{cases}$$ $$v_A(k)=v_B(k) =?$$ $$\begin{aligned} v_A(t)&=\dot{x}_A=\dxt{x_A}\\ &=\dt(6 .00t^2 + 3 .00t + 2.00)\\ &=12t+3 \end{aligned}$$ $$ \begin{aligned} v_B(t)&=\int_0^ta\dd t\\ &=\int(-8.00t)\dd t\\ &=-4t^2+C\\ &=-4t^2+15 \ (\becau..

$$x(16)-x(0) =?$$ $$\begin{aligned} \Ans &= \int_0^{16}v\dd t\\ &= \int_0^2v\dd t+\int_2^{10}v\dd t+\int_{10}^{12}v\dd t+\int_{12}^{16}v\dd t\\ &=\frac{1}{2} \cdot 2 \cdot 8 + 8 \cdot 8+\frac{1}{2} \cdot 2 \cdot (8+4)+4\cdot4\\ &= 100\ut{m} \end{aligned}$$

$$v(40) =?$$ $$\begin{aligned} \Ans &= \int_0^{\frac{40}{1000}}a\dd t\\ &= \int_{\frac{10}{1000}}^{\frac{20}{1000}}a\dd t+\int_{\frac{20}{1000}}^{\frac{30}{1000}}a\dd t+\int_{\frac{30}{1000}}^{\frac{40}{1000}}a\dd t\\ &=\frac{1}{2}\cdot0.01\cdot100+\frac{1}{2}\cdot0.01\cdot(100+400)+\frac{1}{2}\cdot0.01\cdot400\\ &=5\ut{m/s} \end{aligned}$$

$$\begin{cases} A = \text{No Helmat}\\ B = \text{Helmat}\\ \end{cases}$$ $$\abs{\max v_A -\max v_B} =?$$ $$ \begin{aligned} \Ans =&\max v_A -\max v_B \ (\because v_A > v_B)\\ =&v_A(0.007) - v_B(0.007) \ (\because a_A,a_B>0)\\ =&\int_0^{0.007}a_A\dd t-\int_0^{0.007}a_B\dd t\\ =&\(\int_0^{0.002}a_A\dd t+\int_{0.002}^{0.004}a_A\dd t+\int_{0.004}^{0.006}a_A\dd t+\int_{0.006}^{0.007}a_A\dd t\)\\ &-..

(풀이자주:문제에 있는 $v_s=8.0\ut{m}$는 오타로 추정됩니다. 아마 (m/s)를 의도한 것으로 보입니다.) $$\begin{cases} v(0)=0\\ v(10)=2.0\ut{m/s}\\ v(50)=4.0\ut{m/s}\\ \end{cases}$$ (a)$\Delta x_{0\to50}=?$ $$ \begin{aligned} x(50)-x(0)&=\int^{\frac{50}{1000}}_{0}v\dd t\\ &=\int_{0}^{\frac{10}{1000}}v\dd t+\int_{\frac{10}{1000}}^{\frac{50}{1000}}v\dd t\\ &=\frac{1}{2}\cdot \frac{10}{1000} \cdot 2 + \frac{1}{2} \cdot \frac{40}{1000..