10판/4. 2차원운동과 3차원운동

4-10 할리데이 10판 솔루션 일반물리학

짱세디럭스 2020. 6. 27. 22:03
r=5.00t[m]i^+(et+ft2)[m]j^\vec r = 5.00t\ut{m}\hat i +(et+ft^2)\ut{m}\hat j {θ0=35°θ12=0\begin{cases} \theta_0 &= 35\degree\\ \theta_{12} &= 0\\ \end{cases} v= ⁣dr ⁣dt=5.00[m/s]i^+(e+2ft)[m/s]j^\begin{aligned} \vec v =&\dxt{\vec r}\\ =& 5.00\ut{m/s}\hat i +(e+2ft)\ut{m/s}\hat j \end{aligned}
(a) v0=5.00[m/s]i^+(e+2f0)[m/s]j^=5.00[m/s]i^+e[m/s]j^\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} θ0=tan1(e5.00)=35°\begin{aligned} \theta_0 =& \tan^{-1}\(\frac{e}{5.00}\)\\ =& 35\degree\\ \end{aligned} e=5tan35°[m/s]3.50103769105[m/s]3.50[m/s]\begin{aligned} e =& 5\tan35\degree\ut{m/s}\\ \approx&3.50103769105\ut{m/s}\\ \approx&3.50\ut{m/s}\\ \end{aligned}
(b) v12=5.00[m/s]i^+(5tan35°+2f12)[m/s]j^=5.00[m/s]i^+(5tan35°+24f)[m/s]j^\begin{aligned} \vec v_{12} =&5.00\ut{m/s}\hat i +(5\tan35\degree+2f\cdot 12)\ut{m/s}\hat j\\ =&5.00\ut{m/s}\hat i +(5\tan35\degree+24f)\ut{m/s}\hat j\\ \end{aligned} θ12=tan1(5tan35°+24f5.00)=0\theta_{12} = \tan^{-1}\(\frac{5\tan35\degree+24f}{5.00}\)=0 5tan35°+24f=0\therefore 5\tan35\degree+24f=0 f=5tan35°24[m/s]0.14587657046[m/s]0.146[m/s]\begin{aligned} f =& -\frac{5\tan35\degree}{24}\ut{m/s}\\ \approx&-0.14587657046\ut{m/s}\\ \approx&-0.146\ut{m/s}\\ \end{aligned}