Chapter 8.7, Problem 20E

a.

To determine

To find: the parametric equations that represent the descent path of aircraft.

Answer to Problem 20E

  $\begin{array}{c}x=157.1t\\ y=-157.1t-4.9t^2\end{array}$

Explanation of Solution

Given information:

Initial speed $\left|\overrightarrow{v}\right|=80\text{ km/h}$

  $\theta =45°$

Calculation:

First convert the velocity into meter per second as follows:

  $\begin{array}{c}\left|\overrightarrow{v}\right|=80\text{ km/h }\left(\frac{1000\text{ m}}{1\text{ km}}\right)\left(\frac{1\text{ hour}}{3600\text{ s}}\right)\\ =222.2\text{ meter per second}\end{array}$

The angle of flight is below the horizontal i.e. counterclockwise to $+x$ -axis, so it should be either $\left(360°-\theta \right)$ or $\left(-\theta \right)$ .

Now solve the horizontal and vertical components as shown below.

  $\begin{array}{c}x=t\left|\overrightarrow{v}\right|\cos \theta \\ =t\left(222.2\right)\cos \left(-45°\right)\left[\text{Put the value of }\left|\overrightarrow{v}\right|\text{ and }\theta \right]\\ =157.1t\end{array}$

  $\begin{array}{c}y=t\left|\overrightarrow{v}\right|\sin \theta -\frac{1}{2}g{t}^2\\ =t\left(222.2\right)\sin \left(-45°\right)-\frac{1}{2}\left(9.8\right)t^2\left[\text{Put the value of }\left|\overrightarrow{v}\right|,g\text{ and }\theta \right]\\ =-157.1t-4.9t^2\end{array}$

Thus the parametric equations are $x=157.1t$ and $y=-157.1t-4.9t^2$ .

b.

To determine

To find: the distance descended by the aircraft for the given time.

Answer to Problem 20E

4233.4 meters

Explanation of Solution

Given information:

Parametric equations evaluated in part a. are shown below.

  $\begin{array}{c}x=157.1t\\ y=-157.1t-4.9t^2\end{array}$

  $t=2.5\text{ seconds}$

Calculation:

Substitute the value of $t$ in the parametric equation for $y$ and solve $t$ as shown below.

  $\begin{array}{c}y=-157.1t-4.9t^2\\ =-157.1\left(2.5\right)-4.9{\left(2.5\right)}^2\left[\text{Put the value of }t\right]\\ =-423.4\text{ m}\end{array}$

Thus the distance descended by the aircraft for the given time is 423.4 m.

c.

To determine

To find: the average rate for which the plane will lose altitude.

Answer to Problem 20E

608.4 km/h

Explanation of Solution

Given information:

  $\begin{array}{c}t=2.5\text{ s}\\ y=423.4\text{ m}\end{array}$

Calculation:

The average rate of change of vertical position is the average vertical speed, which can be evaluated as follows:

  $\begin{array}{c}\overrightarrow{v_y}=\frac{y}{t}\\ =\frac{423.4\text{ m}}{2.5\text{ s}}\left[\text{Put the value of }y\text{ and }t\right]\\ =169\frac{\text{m}}{\text{s}}\left(\frac{1\text{ km}}{1000\text{ m}}\right)\left(\frac{3600\text{ s}}{1\text{ h}}\right)\left[\text{Convert into km/h}\right]\\ =608.4\text{ km/h}\end{array}$

Thus the average rate is 608.4 km/h.