From f97a59c9f5a5d13d166c6307081e0a6fcc006a69 Mon Sep 17 00:00:00 2001
From: "W. Trevor King" <wking@drexel.edu>
Date: Fri, 21 Jan 2011 13:36:31 -0500
Subject: [PATCH] Cleaner solution for S&J 8, prob. 2.61.

---
 .../Serway_and_Jewett_8/problem02.61.tex      | 31 +++++++++----------
 1 file changed, 14 insertions(+), 17 deletions(-)

diff --git a/latex/problems/Serway_and_Jewett_8/problem02.61.tex b/latex/problems/Serway_and_Jewett_8/problem02.61.tex
index 13c3c3e..8dd8976 100644
--- a/latex/problems/Serway_and_Jewett_8/problem02.61.tex
+++ b/latex/problems/Serway_and_Jewett_8/problem02.61.tex
@@ -52,24 +52,15 @@ yaxis("$x\U{m}$", 0, LeftTicks);
 \Part{a}
 Kathy overtakes Stan when their positions match.
 \begin{align}
-  x_k = \frac{1}{2}a_k (t-\Delta_t)^2 &= x_s = \frac{1}{2}a_s t^2 \\
-  \frac{a_k}{a_s}(t^2 - 2\Delta_t t + \Delta_t^2) &= t^2 \\
-  \p({\frac{a_k}{a_s} - 1})t^2
-    - 2\Delta_t\frac{a_k}{a_s} t
-    + \Delta_t^2\frac{a_k}{a_s} &= 0 \\
-  a t^2 + b t + c &= 0 \\
-  a &= \p({\frac{a_k}{a_s} - 1}) = \p({\frac{4.90}{3.50} - 1}) = 0.400 \\
-  b &= - 2\Delta_t\frac{a_k}{a_s} = - 2\cdot(1.00\U{s})\frac{4.90}{3.50}
-    = -2.80\U{s} \\
-  c &= \Delta_t^2\frac{a_k}{a_s} = (1.00\U{s})^2\frac{4.90}{3.50}
-    = 1.40\U{s$^2$}
+  x_k = \frac{1}{2}a_k (t_p-\Delta_t)^2 &= x_s = \frac{1}{2}a_s t_p^2 \\
+  a_k (t_p-\Delta_t)^2 &= a_s t^2 \\
+  \sqrt{a_k} (t_p-\Delta_t) &= \pm\sqrt{a_s} t_p \\
+  t_p (\sqrt{a_k} \mp \sqrt{a_s}) &= \sqrt{a_k}\Delta_t \\
+  t_p &= \frac{\sqrt{a_k}\Delta_t}{\sqrt{a_k} \mp \sqrt{a_s}}
+     = \frac{\sqrt{4.90\U{m/s$^2$}} \cdot 1.00\U{s}}
+            {\sqrt{4.90\U{m/s$^2$}} \mp \sqrt{3.50\U{m/s$^2$}}}
+     = 6.46 \text{ or } 0.542 \U{s}
 \end{align}
-Solving with the quadratic equation
-\begin{align}
-  t &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
-    = (6.46, 0.542)\U{s}
-\end{align}
-
 The smaller time occurs before Kathy starts, where the quadratic
 equation for $x_k(t)$ does not hold ($x_k(t < \Delta(t))=0$, as Kathy
 is sitting at the starting line).  Therefore, the Kathy passes Stan at
@@ -83,6 +74,12 @@ Kathy travels the same distance as Stan
     = \frac{1}{2} (3.50\U{m/s$^2$}) \cdot (6.46\U{s})^2
     = \ans{73.0\U{m}}
 \end{equation}
+Symbolically,
+\begin{equation}
+  x_s(t_p) = \frac{1}{2}a_s
+             \p({ \frac{\sqrt{a_k}\Delta_t}{\sqrt{a_k} \mp \sqrt{a_s}} })^2
+    = \frac{a_s a_k \Delta_t^2}{2(\sqrt{a_k} \mp \sqrt{a_s})^2}
+\end{equation}
 
 \Part{c}
 The velocities at the passing point are
-- 
2.26.2