Chapter P.4, Problem 47E

Solution

The horizontal distance the airplane will fly to reach an altitude of 12000 ft above the takeoff point.

  $32000\text{ ft}$

Given:

A commercial jet airplane climbs at takeoff with slope $m=\frac{3}{8}$ .

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:

Let the horizontal distance that airplane will fly be x .

Let the take off point be at origin $\left(0,0\right)$ , then the terminal point at an altitude of 12000 ft would be $\left(x,12000\right)$ .

Since the airplane climbs with slope $m=\frac{3}{8}$ , use the slope formula with points $\left(0,0\right)$ and $\left(x,12000\right)$ , and simplify further as shown below

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ \frac{3}{8}=\frac{12000-0}{x-0}\\ \frac{3}{8}=\frac{12000}{x}\\ x=\frac{12000\times 8}{3}\\ x=32000\end{array}$

Thus, the airplane must fly $32000\text{ ft}$ in the horizontal direction.