Chapter 8.6, Problem 33E

Solution

To find: The vector $v$ that represents the plane’s velocity relative to the point of takeoff.

  $v=-195.01i-7.07j+68.40k$

Given information:

Airplane headed west and is climbing at an angle of $20°$ .

Airspeed is 200 mph.

Wind is out of the northeast at 10 mph.

Formula used:

The formula to find the velocity relative to the ground is $v=a+w$ , where $a$ is the plane’s air velocity and $w$ is the wind velocity.

Calculation:

Let $i$ point west $j$ point north and $k$ point up. The plane’s air velocity is

  $a=〈200\cos 160°,0,200\sin 160°〉$

The wind velocity, which is pointing northeast is:

  $w=〈10\cos 220°,10\sin 220°,0〉$

The velocity relative to the ground is $v=a+w$ , so

  $\begin{array}{c}v\approx 〈-187.94,0,68.40〉+〈-7.07,-7.07,0〉\\ \approx 〈-195.01,-7.07,68.40〉\\ \approx -195.01i-7.07j+68.40k\end{array}$

Hence, the velocity the velocity of the plane relative to the point of takeoff is $-195.01i-7.07j+68.40k$