Chapter 11.5, Problem 40PPE

To determine

To find the average speed during first half of the given race.

Answer to Problem 40PPE

7.54 miles per hour.

Explanation of Solution

Given:

Time taken to complete the 10-mile race is 94 minutes.

Average speed during first half is 2 mile per hour greater than the average speed in second half.

Concept Used:

1. One hour is equals to 60 minutes
2. $\text{Time}=\frac{\text{distance}}{\text{Speed}}$

Calculation:

To find the average speed during first half of the given race.

First convert the time taken to complete the race from minutes to hours by dividing it by 60, since one hour is equals to 60 minutes, as

  $\begin{array}{c}94\text{ minutes}=\frac{94}{60}\\ =\frac{47}{30}\text{ hours}\end{array}$

Now, since average speed during first half is 2 mile per hour greater than the average speed in second half thus if average speed in second half is considered to be x mile per hour, then the average speed in first half will be

  $\left(x+2\right)\text{ miles per hour}$

Now since the total time taken is $\frac{47}{30}\text{ hours}$ so according to the given information form the equation in x using the formula

  $\text{Time}=\frac{\text{distance}}{\text{Speed}}$

As,

  $\begin{array}{c}\text{Total time taken=}\left(\begin{array}{l}\text{Total time taken}\\ \text{to complete first}\\ \text{half}\end{array}\right)+\left(\begin{array}{l}\text{Total time taken}\\ \text{to complete }\\ \text{second half}\end{array}\right)\\ \frac{47}{30}=\frac{5}{x+2}+\frac{5}{x}\\ \frac{47}{30}=\frac{5x+5\left(x+2\right)}{x\left(x+2\right)}\\ \frac{47}{30}=\frac{5x+5x+10}{x\left(x+2\right)}\\ 47x\left(x+2\right)=30\left(10x+10\right)\\ 47x^2+94x=300x+300\\ 47x^2-206x-300=0\end{array}$

Now, solve the equation $47x^2-206x-300=0$ , as

  $\begin{array}{c}47x^2-206x-300=0\\ \Rightarrow x=\frac{206\pm \sqrt{{206}^2-4\left(47\right)\left(-300\right)}}{2\times 47}\\ \Rightarrow x=\frac{206\pm \sqrt{4\left(10609+14100\right)}}{2\times 47}\\ \Rightarrow x=\frac{103\pm \sqrt{24709}}{47}\\ \text{Since, x can't be negative}\\ \Rightarrow x=\frac{103+\sqrt{24709}}{47}\\ \approx 5.54\text{ miles per hour}\end{array}$

And since, average speed during first half is 2 mile per hour greater than the average speed in second half.

Thus, average speed in first half is

  $5.54+2=7.54\text{ miles per hour}$ .