Chapter 1.11, Problem 40E

(A)

To determine

To find:

The equation given that the statement

Answer to Problem 40E

The equation is $F=\frac{kw{s}^2}{r}$

Explanation of Solution

Given:

  $F$ varies jointly as $w$ or $F$ is jointly proportional to $w\text{and}{s}^2$

Concept used:

If the quantities $x$ , $y$ and $z$ are related by an equation

  $z=kxy$

Where $k$ is nonzero constant

That is $z$ varies jointly as $x$ or z isjointly proportional to $x\text{and}y$

The constant $k$ is called constant proportionality

Calculation:

If the quantities $F,w$ and $s^2$ and inversely as r are related by an equation

  $F\alpha \frac{w{s}^2}{r}$

  $F=\frac{kw{s}^2}{r}$

For some constant $k\ne 0$

That is $F$ varies jointly as $w$ or $F$ is jointly proportional to $w\text{and}{s}^2$

The constant $k$ is called constant proportionality

(B)

To determine

To find:

The equation of the value given that the statement

Answer to Problem 40E

The speed of second car $s=48mi/h$

Explanation of Solution

Given:

  $F$ varies jointly as $w$ or $F$ is jointly proportional to $w\text{and}{s}^2$

Concept used:

If the quantities $x$ , $y$ and $z$ are related by an equation

  $z=kxy$

Where $k$ is nonzero constant

That is $z$ varies jointly as $x$ or z is jointly proportional to $x\text{and}y$

The constant $k$ is called constant proportionality

Calculation:

The equation of the set is $w=1600lb$ and $s=60mi/h$

The force

  $F_1=\frac{k\left(1600\right){\left(60\right)}^2}{r}$

The second car inside the curve set $w=2500lb$ in the equation

  $F_2=\frac{k\left(2500\right){\left(s\right)}^2}{r}$

Since the force is the same in both cases solve the equation

  $\begin{array}{l}{F}_1=F_2\\ \frac{k\left(1600\right){\left(60\right)}^2}{r}=\frac{k\left(2500\right){\left(s\right)}^2}{r}\\ \left(1600\right)\left(3600\right)=2500{\left(s\right)}^2\\ {\left(s\right)}^2=\frac{\left(1600\right)\left(3600\right)}{2500}\\ {\left(s\right)}^2=2304\\ s=\sqrt{2304}\\ s=48mi/h\end{array}$

The speed of second car $s=48mi/h$