Chapter 12.8, Problem 10WE

To determine

To calculate: How does $t$ changes when $F$ remains constant, $s$ is tripled, and $r$ is halved.

In the formula $t^2=\frac{Fs}{r}$ .

Answer to Problem 10WE

The value of $t$ is $\sqrt{6}$ times as grater as the original value of $t$ , Then to more simplified $\sqrt{6}$ gives $\sqrt{6}\approx 2.4$ .

Explanation of Solution

Given information:

How does $t$ changes when $F$ remains constant, $s$ is tripled, and $r$ is halved.

In the formula $t^2=\frac{Fs}{r}$ .

Calculation:

Consider does $t$ changes when $F$ remains constant, $s$ is tripled, and $r$ is halved.

In the formula $t^2=\frac{Fs}{r}$ .

As $y$ varies directly as $x$ , then the variation is direct and its equation is in the form $y=kx$ .

The variation as:

  $t^2=\frac{Fs}{r}$

Now find $t$ by take the square root on both sides:

  $\begin{array}{c}{t}^2=\frac{Fs}{r}\\ \sqrt{t^2}=\pm \sqrt{\frac{Fs}{r}}\\ t=\pm \sqrt{\frac{Fs}{r}}\end{array}$

Apply $s=3s,r=0.5r$ in the above result:

  $\begin{array}{c}t=\pm \sqrt{\frac{Fs}{r}}\\ t=\pm \sqrt{\frac{F\left(3s\right)}{0.5r}}\\ t=\pm \sqrt{\frac{3Fs}{0.5r}}\\ t=\pm \sqrt{\frac{6Fs}{r}}\end{array}$

Simplified further to get the value of $t$:

  $\begin{array}{l}t=\pm \sqrt{\frac{6Fs}{r}}\\ t=\pm \sqrt{\frac{6\left(Fs\right)}{r}}\\ t=\pm \sqrt{\frac{Fs}{r}\cdot 6}\\ t=\pm \sqrt{\frac{Fs}{r}}\cdot \sqrt{6}\end{array}$

The value of $t$ is:

  $t=\pm \sqrt{\frac{Fs}{r}}\cdot \sqrt{6}$

Then, the value of $t$ is $\sqrt{6}$ times as grater as the original value of $t$ .

Then to more simplified $\sqrt{6}$ gives $\sqrt{6}\approx 2.4$ .

Thus, the value of $t$ is $\sqrt{6}\approx 2.4$ .