Chapter 8.6, Problem 64E

Solution

a.

To show that the plane’s compass bearing is ${83.34}^{\circ }$ .

Given information:

The velocity of the plane relative to the ground is given as $v=193.88i+22.63j+125k$ .

Calculation:

The velocity vector ignoring the upward component is $v=193.88i+22.63j$ .

The angle made the by plane with $x$ -axis is ${\tan }^{-1}\left(\frac{22.63}{193.88}\right)={6.6575}^{\circ }\approx {6.66}^{\circ }$ .

So, the compass bearing of the plane ( $y$ -axis represents the north direction) is ${90}^{\circ }-{6.66}^{\circ }={83.34}^{\circ }$ .

b.

To show that the plane’s downrange speed is $195.2$ mph.

Given information:

The velocity of the plane relative to the ground is given as $v=193.88i+22.63j+125k$ .

Calculation:

The velocity vector ignoring the upward component is $v=193.88i+22.63j$ .

The magnitude of the vector $v=193.88i+22.63j$ gives the plane’s downrange speed.

So, the plane’s downrange speed is $\sqrt{{193.88}^2+{22.63}^2}\approx 195.2$ mph.

c.

To show that the plane is climbing at an angle

Given information:

The velocity of the plane relative to the ground is given as $v=193.88i+22.63j+125k$ .

Calculation:

From part (b), the magnitude of the downrange velocity is 195.2 mph. The upward velocity of the plane is 125 mph.

So, the angle at which the plane is climbing is ${\tan }^{-1}\left(\frac{125}{195.2}\right)={32.6342}^{\circ }\approx {32.63}^{\circ }$ .

That means, the plane is climbing at angle ${32.63}^{\circ }$ .

d.

To show that the plane’s overall speed is $231.8$ mph.

Given information:

The velocity of the plane relative to the ground is given as $v=193.88i+22.63j+125k$ .

Calculation:

The overall speed of the plane is the magnitude of the velocity vector $v=193.88i+22.63j+125k$ .

So, the overall speed is $\sqrt{{193.88}^2+{22.63}^2+{125}^2}=231.7899\approx 231.8$ mph.