\begin{problem*}{3.36}
Three displacement vectors of a croquet ball are shown in Figure
P3.36, where $|\vect{A}| = 20.0\U{units}$, $|\vect{B}| =
40.0\U{units}$, and $|\vect{C}| = 30.0\U{units}$.  Find \Part{a} the
resultant in unit-vector notation and \Part{b} the magnitude and
direction of the resultant displacement.
\begin{center}
\begin{asy}
import graph;
import Mechanics;

real u = 0.06cm;

pair a = 20u * dir(90);
pair b = 40u * dir(45);
pair c = 30u * dir(-45);

Vector A = Vector((0,0), mag=length(a), dir=degrees(a),
    L=Label("$\vect{A}$", position=EndPoint, align=E));
A.draw();
Vector B = Vector((0,0), mag=length(b), dir=degrees(b),
    L=Label("$\vect{B}$", position=EndPoint, align=RightSide));
B.draw();
Vector C = Vector((0,0), mag=length(c), dir=degrees(c),
    L=Label("$\vect{C}$", position=EndPoint, align=LeftSide));
C.draw();

xaxis("$x$");
yaxis("$y$");
\end{asy}
\end{center}
\end{problem*}

\begin{solution}
\Part{a}
\begin{center}
\begin{asy}
import graph;
import Mechanics;

real u = 0.06cm;

pair a = 20u * dir(90);
pair b = 40u * dir(45);
pair c = 30u * dir(-45);
pair r = a + b + c;

Vector A = Vector((0,0), mag=length(a), dir=degrees(a), "$\vect{A}$");
A.draw();
Vector B = Vector(a, mag=length(b), dir=degrees(b), "$\vect{B}$");
B.draw();
Vector C = Vector(a+b, mag=length(c), dir=degrees(c), "$\vect{C}$");
C.draw();
Vector R = Vector((0,0), mag=length(r), dir=degrees(r), "$\vect{r}$");
R.draw();

xaxis("$x$");
yaxis("$y$");
\end{asy}
\end{center}
The resultant displacement is
\begin{align}
  \vect{r} &= \vect{A} + \vect{B} + \vect{C} \\
    &= \{  20.0\jhat
        + 40.0[\cos(45\dg)\ihat + \sin(45\dg)\jhat]
        + 30.0[\cos(-45\dg)\ihat + \sin(-45\dg)\jhat]\}\U{units} \\
    &= (49.5\ihat + 27.1\jhat)\U{units}
\end{align}

\Part{b}
\begin{align}
  |\vect{r}| &= \sqrt{\vect{r}_x^2 + \vect{r}_y^2}
    = \sqrt{49.5^2 + 27.1^2}\U{units}
    = \ans{56.4\U{units}} \\
  \theta_\vect{r} &= \arctan\p({\frac{27.1}{49.5}}) = \ans{28.7\dg}
\end{align}
\end{solution}
