Chapter 11.1, Problem 44PPS

To determine

To Find: The distance when $125$ pound person sits from the center to balance the seesaw.

Answer to Problem 44PPS

  $1.699m$

Explanation of Solution

Given: When two people are balanced on a seesaw their distances from the center of the seesaw is inversely proportional to their weight. If a $118$ pounds person sit $1.8m$ from the center of the seesaw.

The inverse variation equation

  $b=\frac{k}{a}$

Let $b$ be the weight of the person and $a$ be the distance from the center

Then

  $\begin{array}{l}118=\frac{k}{1.8}\\ k=118(1.8)\\ k=212.4\end{array}$

Now

Finding the distance when $125$ pound person sits from the center to balance the seesaw i.e.

  $\begin{array}{l}125=\frac{212.4}{a}\\ 125a=212.4\\ a=1.699m\end{array}$

Hence, the distance when $125$ pound person sits from the center to balance the seesaw is $1.699m$