Chapter 12.8, Problem 6WE

To determine

To calculate: the value of, $v$ , where $r=4.5,a=10$ .

If $a$ varies directly as $r$ and inversely as square of $v$ . If $a=15$ when $r=27$ and $v=8$ .

Answer to Problem 6WE

The value of $v$ is $\pm 4$ .

Explanation of Solution

Given information:

The value of, $v$ , where $r=4.5,a=10$ .

If $a$ varies directly as $r$ and inversely as square of $v$ . If $a=15$ when $r=27$ and $v=8$ .

Calculation:

Consider the value of, $v$ , where $r=4.5,a=10$ .

If $a$ varies directly as $r$ and inversely as square of $v$ . If $a=15$ when $r=27$ and $v=8$ .

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

When it varies inversely as $x$ , then the variation in inverse and its equation is in the form of

  $y=\frac{k}{x}$

Where $k$ is the constant of variation.

Recall that the $a$ ,varies inversely as the square of $v$ and directly as $r$, then:

  $a=k\frac{r}{v^2}$

Here , If $r=27$ when $v=8$ and the value of $a=15$ .

  $\begin{array}{c}a=k\frac{r}{v^2}\\ 15=k\frac{27}{8^2}\\ 15=\frac{27k}{64}\\ 15\times 54=27k\end{array}$

From here the value of $k$ is:

  $\begin{array}{c}15\times 54=27k\\ 960=27k\\ \frac{960}{27}=k\\ \frac{320}{9}=k\end{array}$

Thus the equation of the variation is :

  $\begin{array}{l}a=k\frac{r}{v^2}\\ a=\frac{\frac{320}{9}r}{v^2}\\ a=\frac{320r}{90v^2}\end{array}$

Now to find the value of $v$ when $r=4.5$ and $a=10$ .

Now apply these values in the above expression:

  $\begin{array}{c}a=\frac{320r}{90v^2}\\ 10=\frac{320\times 4.5}{9v^2}\\ 10=\frac{1440}{9v^2}\\ 10\times 9v^2=1440\end{array}$

Then the value of $h$ is:

  $\begin{array}{c}10\times 9v^2=1440\\ 90v^2=1440\\ v^2=\frac{1440}{90}\\ v^2=16\end{array}$

That is:

  $\begin{array}{c}{v}^2=16\\ v=\pm \sqrt{16}\\ v=\pm 4\end{array}$

Thus, the value of $v$ is $\pm 4$ . $$