Use non-breaking space (~) between 'Figure' and the figure number.
[course.git] / latex / problems / Serway_and_Jewett_8 / problem03.28.tex
index 3fbb4237e2748342cae25d7b004b8c066de38203..ac7603eae9634aa6ddf2554441ac2f787f8ffe71 100644 (file)
@@ -8,4 +8,45 @@ magnitude of the football's resultant displacement?
 \end{problem*}
 
 \begin{solution}
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real u = 0.1cm;
+
+pair drop = (-10u, 0);
+pair scramble = (0, 15.0u);
+pair pass = (50u, 0);
+pair result = drop+scramble+pass;
+
+dot((0,0));  // initial ball position
+draw((0,-5u)--(0, 20u), dashed);  // line of scrimmage
+
+Vector vDrop = Vector((0,0), mag=length(drop), dir=degrees(drop), "drop");
+vDrop.draw();
+Vector vScramble = Vector(
+    drop, mag=length(scramble), dir=degrees(scramble),
+    L=Label("scramble", align=W));
+vScramble.draw();
+Vector vPass = Vector(drop+scramble, mag=length(pass), dir=degrees(pass),
+    "pass");
+vPass.draw();
+Vector vResult = Vector((0,0), mag=length(result), dir=degrees(result),
+    "result");
+vResult.draw();
+\end{asy}
+\end{center}
+
+The resultant displacement is
+\begin{equation}
+  \vect{r} = \vect{d} + \vect{s} + \vect{p}
+    = (-10.0\ihat + 15.0\jhat + 50.0\ihat)\U{yards}
+    = (40.0\ihat + 15.0\jhat)\U{yards}
+\end{equation}
+Which has a magnitude
+\begin{equation}
+  |\vect{r}| = \sqrt{\vect{r}_x^2 + \vect{r}_y^2}
+    = \sqrt{40.0^2 + 15.0^2}\U{yards}
+    = \ans{42.7\U{yards}}
+\end{equation}
 \end{solution}