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$$
\begin{array}{l}{12-221 . \text { Two boats leave the pier } P \text { at the same time and }} \\ {\text { travel in the directions shown. If } v_{A}=40 \mathrm{ft} / \mathrm{s} \text { and } v_{B}=} \\ {30 \mathrm{ft} / \mathrm{s}, \text { determine the velocity of boat } A \text { relative to boat } B \text { . }} \\ {\text { How long after leaving the pier will the boats be } 1500 \mathrm{ft} \text { apart? }}\end{array}
$$

$$
v_{A}=40ft/ s
$$

$$
v_{B}=30 f t / s
$$

$$
v_{A/B}=? ?
$$

$$
s=1500f t
$$

$$
V_{A}=V_{B}+V_{A/B}
$$

\( {∴ 40 \sin 30^{\circ}i+40 \cos 30^{\circ}j=30 \cos 45^{\circ} i+30 \sin 45^{\circ}j+v_{A/B}} \)

\( {∴ V_{\mathrm{A} / \mathrm{B}}=(-1.213 i+13.43 j) \mathrm{ft} / \mathrm{s}}\ \)

$$
V_{A/B}=\sqrt{(-1.213)^{2}+(13.43)^{2}} \simeq 13.5 f t / s
$$

$$
\theta=\tan^{-1}  \frac{13.43}{1.213}=84.8^{\circ}
$$

$$
v=\frac{s}{t} \rightarrow t=\frac{s}{v}
$$

\(∴ t=\frac{1500}{V_{A/B}}=\frac{1500}{13 \cdot 5} \simeq 111.27s \)

$$
\simeq 1.85 \mathrm{min}
$$

$$
\begin{array}{l}{12-219 . \text { At the instant shown, cars } A \text { and } B \text { are traveling at }} \\ {\text { speeds of } 55 \mathrm{mi} / \mathrm{h} \text { and } 40 \mathrm{mi} / \mathrm{h} \text { , respectively. If } B \text { is }} \\ {\text { increasing its speed by } 1200 \mathrm{mi} / \mathrm{h}^{2} \text { , while } A \text { maintains a }} \\ {\text { constant speed, determine the velocity and acceleration of } B} \\ {\text { with respect to } A \text { . Car } B \text { moves along a curve having a }} \\ {\text { radius of curvature of } 0.5 \mathrm{mi} .}\end{array}
$$

$$
v_{A}=55 \mathrm{mi/h}
$$

$$
v_{B}=40 \text { mi/h }
$$

$$
a_{B}=1200 \mathrm{mi} / \mathrm{h}^{2}
$$

$$
r=0.5  mi
$$

$$
v_{A}=(-55 i) m{i} / h
$$

\(∴ v_{B}=-40 \cos 30^{\circ} i+ 40 \sin 30^{\circ} j \)

\(∴ v_{B}=(-34.64 i+20 j) mi / h \)

\(∵ v_{B}=V_{A}+V_{B/A} \)

\(∴v_{B/A}-v_{B}-v_{A}=(-34.64 i+20j)-(-55i) \)

$$
=(20.36 i+20 j) \mathrm{mi} / \mathrm{h}
$$

$$
V_{ B / A}=\sqrt{(20.36)^{2}+20^{2}}=28.5 \mathrm{mi} / \mathrm{h}
$$

$$
\theta=\tan ^{-1} \frac{20}{20.36}=44.5^{\circ}
$$

$$
v_{A}=55 \mathrm{mi/h}
$$

$$
v_{B}=40 \text { mi/h }
$$

$$
a_{B}=1200 \mathrm{mi} / \mathrm{h}^{2}
$$

$$
r=0.5  mi
$$

$$
\left(a_{B}\right)_{n}=\frac{v_{B}^{2}}{\rho}=\frac{40^{2}}{0.5}=3200 \mathrm{mi/h}^{2} \ ,\left(a_{B} \right)_{t}=1200 \mathrm{mi} / \mathrm{h}^{2}
$$

$$
{\left(a_{B}\right)_{n}=3200 \cos 60 i+3200 \sin 60 j}\
$$

$$
\left(a_{B}\right)_{t}=1200 \cos 30 i+1200 \sin 30j
$$

$$
 {a_{B}=(560.77 i+3371.28j) \mathrm{mi} / \mathrm{h}^{2}}\
$$

$$
a_{A}=0 \Rightarrow  a_{B/A}=a_{B}-a_A=a_{B}
$$

$$
=(560.77 i+3371.28 j) mi/ h ^{2}
$$

$$
a_{B / A}=\sqrt{(560.77)^{2}+(3371. 28)^{2}}=3418 mi /h ^{2}
$$

$$
\theta=\tan ^{-1} \frac{3371.28}{560.77}=80.6^{\circ}
$$

 

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