Solution

Engineering Mechanics (Dynamics)

Chapter 12

Exercise 221

Relative velocity can be determined from the equation:

$$\begin{align*} \vec{v}_A=\vec{v}_B+\vec{v}_{A/B} \end{align*}$$

where the known velocities in the reference frame given in the problem are:

$$\begin{align*} \vec{v}_A=v_A\sin30\text{\textdegree} \textbf{i} + v_A\cos30\text{\textdegree} \textbf{j}\\ \vec{v}_B=v_B\cos45\text{\textdegree} \textbf{i} + v_B\sin45\text{\textdegree} \textbf{j} \end{align*}$$

Substituting these into the first relation, with $v_A=40 \:\text{ft}/\text{s}$ and $v_B=30 \:\text{ft}/\text{s}$, we get:

$$\begin{align*} \vec{v}_{A/B}&=\vec{v}_A-\vec{v}_B\\ &=v_A\sin30\text{\textdegree} \textbf{i} + v_A\cos30\text{\textdegree} \textbf{j}-v_B\cos45\text{\textdegree} \textbf{i} - v_B\sin45\text{\textdegree} \textbf{j}\\ &=(v_A\sin30\text{\textdegree} - v_B\cos45\text{\textdegree})\textbf{i}+(v_A\cos30\text{\textdegree}-v_B\sin45\text{\textdegree})\textbf{j}\\ &=(40\sin30\text{\textdegree} - 30\cos45\text{\textdegree})\textbf{i}+(40\cos30\text{\textdegree}-30\sin45\text{\textdegree})\textbf{j}\\ &= (-1.213\textbf{i}+ 13.428\textbf{j}) \:\text{ft}/\text{s} \end{align*}$$

The magnitude of this relative velocity is then:

$$\begin{align*} v_{A/B}=\sqrt{(-1.213)^2+13.428^2}=\boxed{13.5 \:\text{ft}/\text{s}} \end{align*}$$

t=vA/B​d​=13.51500​=111.11s=1.85min
​

$$$$
\begin{align*} t=\dfrac{d}{v_{A/B}}=\dfrac{1500}{13.5}=111.11 \:\text{s}=\boxed{1.85 \:\text{min}} \end{align*}