Chapter 6, Problem 1E

(a)

To determine

To Find:

Solve the given equation for s , the speed of car.

Answer to Problem 1E

  $s=60\sqrt{f}$

Explanation of Solution

Given Information:

The equation

  $\begin{array}{l}d=\frac{s^2}{30f}\\ d=\text{length of skid marks (in feet)}\\ \text{s = speed of car ( in miles per hour)}\\ \text{f = coefficient of friction}\end{array}$

can be used to model the length of the skid marks left by a car when a driver applies the breaks to come to a sudden stop.Suppose a car left skid marks that are 120 feet long.

Calculation:

  $\begin{array}{l}Given\\ d=\frac{s^2}{30f}\\ d=\text{length of skid marks (in feet)=120}\\ \text{s = speed of car ( in miles per hour)}\\ \text{f = coefficient of friction}\\ \text{So, the equation becomes:}\\ 120=\frac{s^2}{30f}\\ 120(30f)=s^2\mathrm{........}[\text{multiply each side by 30f]}\\ s=\pm \sqrt{3600f}\mathrm{.........}[\text{taking square root both sides ]}\\ s=\pm 60\sqrt{f}\\ \text{Since , speed can never be negative}\text{.}\\ \text{Hence , the speed of the car is represented by the equation:}\\ s=60\sqrt{f}\end{array}$

(b)

To determine

To Find:

The speed of car.

Answer to Problem 1E

About 52 miles per hour

Explanation of Solution

Given Information:

Given

f = 0.75

From part (a) , the speed of car is given by:

  $s=60\sqrt{f}$

Calculation:

  $\begin{array}{l}Given\\ s=60\sqrt{f}\\ \text{s = speed of car ( in miles per hour)}\\ \text{f = coefficient of friction = 0}\text{.75}\\ \text{Substitute f = 0}\text{.75 in the equation :}\\ s=60\sqrt{0.75}\\ \text{⇒}s\approx 60(0.8660)\\ \text{⇒}s\approx 51.96\approx 52\\ \text{So, the speed of the car is about 52 miles per hour}\text{.}\end{array}$

(c)

To determine

To Find:

The speed of car.

Answer to Problem 1E

About 63 miles per hour

Explanation of Solution

Given Information:

Given

f = 1.1

From part (a) , the speed of car is given by:

  $s=60\sqrt{f}$

Calculation:

  $\begin{array}{l}Given\\ s=60\sqrt{f}\\ \text{s = speed of car ( in miles per hour)}\\ \text{f = coefficient of friction = 1}\text{.1}\\ \text{Substitute f = 1}\text{.1 in the equation :}\\ s=60\sqrt{1.1}\\ \text{⇒}s\approx 60(1.0488)\\ \text{⇒}s\approx 62.9285\approx 63\\ \text{So, the speed of the car is about 63 miles per hour}\text{.}\end{array}$