Chapter P.4, Problem 48E

Solution

a.

The slope of the section of the highway.

  $m=0.06$

Given:

The highway has a section posted as a $6\%$ grad. This means that for a horizontal change of 100 ft there is a 6-ft vertical change.

Concept Used:

• The slope-intercept form of a line is $y=mx+b$, where m is the slope of the line and b is the y -intercept of the line.
• Slope of the line passing through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}$
• The slope of a line is defined as rise over run

Calculation:

The gradient of the highway represents the slope of the highway.

Since the gradient of the highway is 6%, the slope of the highway is

  $\begin{array}{c}m=6\%\\ =\frac{6}{100}\\ =0.06\end{array}$

Thus, the slope of the section of given highway is $m=0.06$ .

b.

The horizontal distance required to travel to climb 250 ft.

  $\text{About 4167 ft}$

Given:

The highway has a section posted as a $6\%$ grad. This means that for a horizontal change of 100 ft there is a 6-ft vertical change.

Concept Used:

• The slope-intercept form of a line is $y=mx+b$, where m is the slope of the line and b is the y -intercept of the line.
• Slope of the line passing through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}$
• The slope of a line is defined as rise over run

Calculation:

Since it required to travel 100 ft horizontally for a vertical change of 6-ft, the unit rate for the horizontal distance for climbing 1-ft in the given section of highway is $\frac{100}{6}$ .

Hence, in order to climb 250 ft, the horizontal distance required to travel is

  $\begin{array}{c}\frac{100}{6}\cdot 250=\frac{25000}{6}\\ \approx 4167\end{array}$

Thus, the horizontal distance to be traveled is about $4167$ ft.

c.

The vertical change for the next 7 miles.

  $\text{About }2217.6\text{ ft}$

Given:

The highway has a section posted as a $6\%$ grad. This means that for a horizontal change of 100 ft there is a 6-ft vertical change.

Concept Used:

• The slope-intercept form of a line is $y=mx+b$, where m is the slope of the line and b is the y -intercept of the line.
• Slope of the line passing through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}$
• The slope of a line is defined as rise over run

Calculation:

Since there are 5280 ft in a mile, the total number of ft in 7 miles is

  $5280\times 7=36960$

Since it required to travel 100 ft horizontally for a vertical change of 6-ft, the unit rate for the vertical change for each 1-ft horizontal change in the given section of highway is $\frac{6}{100}$ .

Hence, in order to travel 36960 ft horizontally, the vertical change is

  $\begin{array}{c}\frac{6}{100}\cdot 36960=\frac{221760}{100}\\ \approx 2217.6\end{array}$

Thus, the vertical change will be about $2217.6$ ft.