4-10 할리데이 10판 솔루션 일반물리학

$$\vec r = 5.00t\ut{m}\hat i +(et+ft^2)\ut{m}\hat j$$ $$\begin{cases} \theta_0 &= 35\degree\\ \theta_{12} &= 0\\ \end{cases} $$ $$\begin{aligned} \vec v =&\dxt{\vec r}\\ =& 5.00\ut{m/s}\hat i +(e+2ft)\ut{m/s}\hat j \end{aligned}$$ (a) $$\begin{aligned} \vec v_0 =&5.00\ut{m/s}\hat i +(e+2f\cdot 0)\ut{m/s}\hat j\\ =&5.00\ut{m/s}\hat i +e\ut{m/s}\hat j\\ \end{aligned}$$ $$\begin{aligned} \theta_0 =& \tan^{-1}\(\frac{e}{5.00}\)\\ =& 35\degree\\ \end{aligned}$$ $$\begin{aligned} e =& 5\tan35\degree\ut{m/s}\\ \approx&3.50103769105\ut{m/s}\\ \approx&3.50\ut{m/s}\\ \end{aligned}$$ (b) $$\begin{aligned} \vec v_{14} =&5.00\ut{m/s}\hat i +(5\tan35\degree+2f\cdot 14)\ut{m/s}\hat j\\ =&5.00\ut{m/s}\hat i +(5\tan35\degree+28f)\ut{m/s}\hat j\\ \end{aligned}$$ $$\theta_{12} = \tan^{-1}\(\frac{5\tan35\degree+28f}{5.00}\)=0$$ $$\therefore 5\tan35\degree+28f=0$$ $$\begin{aligned} f =& -\frac{5\tan35\degree}{28}\ut{m/s}\\ \approx&-0.12503706039459103\ut{m/s}\\ \approx&-0.125\ut{m/s}\\ \end{aligned}$$