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# CHAPTER 2 Problem 115

<span class="room-tag">In 1889, at Jubbulpore, India, a tug-of-war was finally won after 2 h 41 min, with the winning team displacing the center of the rope 3.7 m. In centimeters per minute, what was the magnitude of the average velocity of that center point during the contest?</span>

<br/>

根據平均速率之定義：
\begin{aligned}
\bar{v} ={{\ell } \over {\Delta t}}
\end{aligned}
<br>

$\ell$為路徑長，其值為3.7m，移動時間為161min，
\begin{aligned}
\bar{v} ={{\ell } \over {\Delta t}}=\frac{3.7\ \mathrm{m}}{161\ \mathrm{min}}=\frac{3.7\times 100\ \mathrm{cm}}{161\ \mathrm{min}} \approx 2.3\ (\mathrm{cm/min})
\end{aligned}
<br>

###### tag: `Fundamentals of Physics`