Solution

Engineering Mechanics (Dynamics)

Chapter 12

Exercise 228

$$\begin{array}{rl}& \\ {\stackrel{⃗}{}}_{}^{}& ={\vec{v}}_B-{\vec{v}}_A\\ & =-v_B\sin 30\text{°}\text{i}+v_B\cos 30\text{°}\text{j}-v_A\text{j}\\ & =(-15\text{i}-14.02\text{j})\text{ft}/\text{s}\end{array}$$

The magnitude of this relative velocity is then:

$$\begin{align*} v_{B/A}=\sqrt{(1.15)^2+(-14.02)^2}=\boxed{20.5 \:\text{m}/\text{s}} \end{align*}$$

The direction of the relative velocity, that is the angle $\theta$ that the vector $\vec{v}_{B/A}$ makes with the horizontal axis can be determined as:

$$\begin{align*} \tan \theta&=\dfrac{14.02}{15}\\ \theta &= \boxed{43.1\text{\textdegree}} \swarrow \end{align*}$$

Answer