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+### Statement
+
+$3.2.15.$ [Insert the problem statement]
+
+### Solution
+
+\documentclass{article}
+\usepackage{graphicx} % Required for inserting images
+\begin{document}
+
+\section{Understanding the motion}
+\[
+\textbf{Solution}
+\]
+
+The elevator descends a distance \(S=400\,\mathrm{m}\) in \(t=40\,\mathrm{s}\).
+During the first half of the descent it accelerates with acceleration \(a\),
+and during the second half it decelerates with the same magnitude.
+
+The total distance traveled is
+
+\[
+S=
+\frac{1}{2}a\left(\frac{t}{2}\right)^2+
+\frac{1}{2}a\left(\frac{t}{2}\right)^2
+=\frac{at^2}{4}.
+\]
+
+Therefore,
+
+\[
+a=\frac{4S}{t^2}
+=\frac{4\cdot400}{40^2}
+=1\,\mathrm{m/s^2}.
+\]
+
+During the first half of the descent, the effective gravitational acceleration
+for the pendulum is
+
+\[
+g_{\mathrm{eff},1}=g-a,
+\]
+
+so its period is
+
+\[
+T_1=2\pi\sqrt{\frac{l}{g-a}}.
+\]
+
+During the second half,
+
+\[
+g_{\mathrm{eff},2}=g+a,
+\]
+
+and therefore
+
+\[
+T_2=2\pi\sqrt{\frac{l}{g+a}}.
+\]
+
+Hence, during the first and second halves of the descent, the numbers of
+oscillations are
+
+\[
+n_1=\frac{t/2}{T_1}
+=\frac{t}{4\pi}\sqrt{\frac{g-a}{l}},
+\]
+
+and
+
+\[
+n_2=\frac{t/2}{T_2}
+=\frac{t}{4\pi}\sqrt{\frac{g+a}{l}}.
+\]
+
+Thus the total number of oscillations during one descent is
+
+\[
+n_{\mathrm{d}}
+=
+\frac{t}{4\pi\sqrt{l}}
+\left(\sqrt{g-a}+\sqrt{g+a}\right).
+\]
+
+If the elevator were not accelerating, the pendulum would make
+
+\[
+n_0=\frac{t}{2\pi}\sqrt{\frac{g}{l}}
+\]
+
+oscillations in the same time.
+
+The number of oscillations lost is therefore
+
+\[
+n_0-n_{\mathrm{d}}
+=
+\frac{t\sqrt{g}}{4\pi\sqrt{l}}
+\left[
+2-\sqrt{1-\frac{a}{g}}
+-\sqrt{1+\frac{a}{g}}
+\right].
+\]
+
+The normal period of the pendulum is
+
+\[
+T_0=2\pi\sqrt{\frac{l}{g}}.
+\]
+
+Therefore, the time lost by the clock during one descent is
+
+\[
+\Delta t=(n_0-n_{\mathrm{d}})T_0.
+\]
+
+After cancellation,
+
+\[
+\boxed{
+\Delta t=
+\frac{t}{2}
+\left[
+2-\sqrt{1-\frac{a}{g}}
+-\sqrt{1+\frac{a}{g}}
+\right]
+}.
+\]
+
+For \(t=40\,\mathrm{s}\), \(a=1\,\mathrm{m/s^2}\), and
+\(g=9.8\,\mathrm{m/s^2}\),
+
+\[
+\Delta t
+=
+20
+\left[
+2-\sqrt{1-\frac{1}{9.8}}
+-\sqrt{1+\frac{1}{9.8}}
+\right]
+\approx 0.054\,\mathrm{s}.
+\]
+
+The same time loss occurs during an ascent, since the two effective
+accelerations \(g-a\) and \(g+a\) simply occur in the opposite order.
+
+In \(5\) hours,
+
+\[
+5\,\mathrm{h}=18000\,\mathrm{s},
+\]
+
+so the number of ascents or descents is
+
+\[
+N=\frac{18000}{40}=450.
+\]
+
+Consequently, the total time lost is
+
+\[
+\Delta T=N\Delta t
+=450(0.054)
+\approx24.3\,\mathrm{s}.
+\]
+
+Therefore,
+
+\[
+\boxed{\Delta T\approx24\,\mathrm{s}}.
+\]
+\end{document}
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]