Chapter 5.4, Problem 53E

a.

To determine

To compare: The stopping distances of a car on the surfaces given in the table.

Answer to Problem 53E

Stopping distance of car on different surfaces is as follows:

Dry asphalt = $90\text{ feet}$

Wet asphalt = $225\text{ feet}$

Snow = $225\text{ feet}$

Ice = $450\text{ feet}$

Explanation of Solution

Given information:

The speed s (in miles per hour) of a car is given by the function,

  $s=\sqrt{30fd}$

where f is the coefficient of friction and d is the stopping distance (in feet).

The following table:

  

The car is travelling $45\text{ miles per hour}$ .

Calculation:

Stopping distance of a car is calculated as:

  $\begin{array}{l}s=\sqrt{30fd}[\text{Given]}\\ s={\left(30fd\right)}^{1/2}[\text{Express in exponent form]}\\ s^2=30fd[\text{Raise each side to second power]}\\ d=\frac{s^2}{30f}[\text{Divide both sides by 30}f]\end{array}$

Stopping distance of a car travelling $45\text{ miles per hour}$ on dry asphalt is calculated as:

  $\begin{array}{l}d=\frac{s^2}{30f}[\text{Calculated]}\\ d=\frac{{45}^2}{30\cdot (0.75)}[\text{Substitute]}\\ d=90\text{ feet }\text{[Simplify]}\end{array}$

Stopping distance of a car travelling $45\text{ miles per hour}$ on wet asphalt is calculated as:

  $\begin{array}{l}d=\frac{s^2}{30f}[\text{Calculated]}\\ d=\frac{{45}^2}{30\cdot (0.30)}[\text{Substitute]}\\ d=225\text{ feet }\text{[Simplify]}\end{array}$

Stopping distance of a car travelling $45\text{ miles per hour}$ on snow is calculated as:

  $\begin{array}{l}d=\frac{s^2}{30f}[\text{Calculated]}\\ d=\frac{{45}^2}{30\cdot (0.30)}[\text{Substitute]}\\ d=225\text{ feet }\text{[Simplify]}\end{array}$

Stopping distance of a car travelling $45\text{ miles per hour}$ on ice is calculated as:

  $\begin{array}{l}d=\frac{s^2}{30f}[\text{Calculated]}\\ d=\frac{{45}^2}{30\cdot (0.15)}[\text{Substitute]}\\ d=450\text{ feet}\text{[Simplify]}\end{array}$

b.

To determine

To calculate: The stopping distance of a car travelling on ice.

Answer to Problem 53E

When travelling on ice at $35\text{ miles per hour}$ , the brake should be applied at $272\text{ feet}$ away from the deer to avoid hitting it.

Explanation of Solution

Given information:

The speed s (in miles per hour) of a car is given by the function,

  $s=\sqrt{30fd}$

where f is the coefficient of friction and d is the stopping distance (in feet).

The following table:

The car is travelling $35\text{ miles per hour}$ .

Calculation:

Stopping distance of a car is calculated as:

  $\begin{array}{l}s=\sqrt{30fd}[\text{Given]}\\ s={\left(30fd\right)}^{1/2}[\text{Express in exponent form]}\\ s^2=30fd[\text{Raise each side to second power]}\\ d=\frac{s^2}{30f}[\text{Divide both sides by 30}f]\end{array}$

Stopping distance of a car travelling $35\text{ miles per hour}$ on ice is calculated as:

  $\begin{array}{l}d=\frac{s^2}{30f}[\text{Calculated]}\\ d=\frac{{35}^2}{30\cdot (0.15)}[\text{Substitute]}\\ d=272\text{ feet }\text{[Simplify]}\end{array}$