Chapter 9.4, Problem 77AYU

a.

To determine

To express: the velocity ${\text{v}}_a$ of the 747 relative to the air and velocity ${\text{v}}_w$ of the jet stream in terms of i and j.

Answer to Problem 77AYU

  $50\sqrt{2}\left(i+j\right)$ .

Explanation of Solution

Given:

Constant speed of Boeing 747 jumbo jet airspeed is $550\text{mi/hr}$

Speed of jet stream is $100\text{mi/hr}$ in the northeasterly direction.

Calculation:

According to the question,

  $\begin{array}{l}{v}_a=550j\\ v_a=100\left(\cos {45}^{\circ }i+i\sin {45}^{\circ }j\right)\\ v_a=100\left(\frac{\sqrt{2}}{2}i+\frac{\sqrt{2}}{2}j\right)\\ =50\sqrt{2}i+50\sqrt{2}j\\ =50\sqrt{2}\left(i+j\right)\end{array}$ .

Hence, the velocity ${\text{v}}_a$ of the 747 relative to the air and velocity ${\text{v}}_w$ of the jet stream in terms of i and j is $50\sqrt{2}\left(i+j\right)$ .

b.

To determine

To find: the velocity of the 747 relatives to the ground.

Answer to Problem 77AYU

  $50\sqrt{2}i+\left(550+50\sqrt{2}\right)j$ .

Explanation of Solution

Given:

Constant speed of Boeing 747 jumbo jet airspeed is $550\text{mi/hr}$

Speed of jet stream is $100\text{mi/hr}$ in the northeasterly direction.

Calculation:

According to the question,

  $\begin{array}{l}{v}_g=v_a+v_w\\ =500j+50\sqrt{2}i+50\sqrt{2}j\\ =50\sqrt{2}i+\left(550+50\sqrt{2}\right)j\end{array}$ .

Hence, the velocity of the 747 relatives to the ground are $50\sqrt{2}i+\left(550+50\sqrt{2}\right)j$ .

c.

To determine

To find: the actual speed and direction of the 747 relatives to the ground.

Answer to Problem 77AYU

  $624.7\text{mgh;}N{6.5}^{\circ }E$ .

Explanation of Solution

Given:

Constant speed of Boeing 747 jumbo jet airspeed is $550\text{mi/hr}$

Speed of jet stream is $100\text{mi/hr}$ in the northeasterly direction.

Calculation:

According to the question,

  $\begin{array}{l}\Vert v_g\Vert =\sqrt{{\left(50\sqrt{2}\right)}^2+{\left(500+50\sqrt{2}\right)}^2}\\ =\sqrt{390281.7459}\\ =624.7\\ \tan \alpha =\frac{\left(500+50\sqrt{2}\right)}{50\sqrt{2}}\\ \therefore \alpha ={83.5}^0\end{array}$ .

Therefore, the direction is ${\left(90-83.5\right)}^{\circ }=N{6.5}^{\circ }E$ .

Hence, the actual speed and direction of the 747 relatives to the ground are $624.7\text{mgh;}N{6.5}^{\circ }E$ .