Chapter 9.1, Problem 60E

(a)

To determine

To find: Expression of wind velocity in component form.

Answer to Problem 60E

The expression of wind velocity in component form is $27.5i+47.6j$.

Explanation of Solution

Given:

The speed of wind is $\text{55}\text{mi/h}$ and direction of  wind is $\text{N}\text{30}°\text{E}$.

Formula used:

If velocity of wind is vector u which makes angle $\theta$ with positive $x$ axis, then wind velocity is

$u=\left|u\right|\cos \theta i+\left|u\right|\sin \theta j$ (1)

Calculation:

The speed of wind is $\text{55}\text{mi/h}$ and direction of wind is $\text{N}\text{30}°\text{E}$.

The angle made by wind velocity with positive $x$ axis is given by

$\begin{array}{c}\theta =\left(90°-30°\right)\\ =60°\end{array}$

Wind velocity vector is shown in Figure (1).

Figure (1)

Substitute $55$ for $\left|u\right|$, $60°$ for $\theta$ in equation (1).

$\begin{array}{c}u=55\cos 60°+55\sin 60°\\ =55\times \frac{1}{2}i+55\times \frac{\sqrt{3}}{2}j\\ =27.5i+47.6j\end{array}$

Thus, component form of wind velocity u is $27.5i+47.6j$.

(b)

To determine

To find: Expression of Jet velocity in component form.

Answer to Problem 60E

Expression of jet velocity in component form relative to air is $540.9i+540.9j$.

Explanation of Solution

Given:

The speed of Jet is $\text{765}\text{mi/h}$ and direction of Jet is $\text{N}45°\text{E}$.

Formula used:

If velocity of Jet is vector v which makes angle $\theta$ with positive $x$ axis, then wind velocity is

$v=\left|v\right|\cos \theta i+\left|v\right|\sin \theta j$ (2)

Calculation: The speed of Jet is $\text{765}\text{mi/h}$ and direction of Jet is $\text{N}45°\text{E}$.

The angle made by Jet velocity vector with positive $x$ axis is given by

$\begin{array}{c}\theta =\left(90°-45°\right)\\ =45°\end{array}$

Jet velocity vector is shown in Figure (2).

Figure (2)

Substitute $765$ for $\left|v\right|$, $45°$ for $\theta$ in equation (2).

$\begin{array}{c}v=765\cos 45°+765\sin 45°\\ =765\times \frac{1}{\sqrt{2}}i+765\times \frac{1}{\sqrt{2}}j\\ =540.9i+540.9j\end{array}$

Thus, component form of Jet velocity v is $540.9i+540.9j$.

(c)

To determine

To find: The true velocity of Jet as a vector.

Answer to Problem 60E

True velocity of jet as a vector is $568.4i+588.5j$.

Explanation of Solution

Given:

The speed of wind is $\text{55}\text{mi/h}$ and direction of  wind is $\text{N}\text{30}°\text{E}$ and the speed of Jet is $\text{765}\text{mi/h}$ and direction of Jet is $\text{N}45°\text{E}$.

Formula used:

If vector u is $u=a_{1}i+a_{2}j$ and vector v is $v=b_{1}i+b_{2}j$,

$u+v=\left(a_1+b_1\right)i+\left(a_2+b_2\right)j$ (3)

Calculation:

If true velocity of jet is vector p , then vector p will be the resultant sum of wind velocity(u) and jet velocity(v).

$p=u+v$

Figure (3) shows the vector u, vector v and vector p on coordinate axis.

Figure (3)

From section(a) wind velocity is $27.5i+47.6j$. and from section (2) jet velocity is $540.9i+540.9j$.

Substitute $27.5$ for $a_1$, $47.6$ for $a_2$, $540.9$ for $b_1$ and $540.9$ for $b_2$ in equation (3).

$\begin{array}{c}u+v=\left(27.5+540.9\right)i+\left(47.6+540.9\right)j\\ =568.4i+588.5j\end{array}$

And vector p is $p=u+v$.

Therefore, $p=568.4i+588.5j$ .

Thus true velocity of jet is $568.4i+588.5j$.

(d)

To determine

To find: True speed and direction of Jet.

Answer to Problem 60E

True speed of jet is $818.18\text{mi/h}$ and direction of jet is $\text{N}\text{43}\text{.88}°\text{E}$.

Explanation of Solution

Given:

The speed of Jet relative to air is $\text{765}\text{mi/h}$ and direction of Jet is $\text{N}45°\text{E}$.

Formula used:

If $p=a_{1}i+a_{2}j$ is a vector in component form then magnitude of vector is,

$\left|p\right|=\sqrt{a_1{}^2+a_2{}^2}$ (3)

Where,

• $a_1$ is the horizontal component.
• $a_2$ is the vertical component.

Calculation:

From Section(c), the true velocity of jet is $568.4i+588.5j$.

Substitute $568.4$ for $a_1$ and $588.5$ for $a_2$ in equation (3).

$\begin{array}{l}\left|p\right|=\sqrt{{\left(568.4\right)}^2+{\left(588.5\right)}^2}\\ =\sqrt{323078.6+346332.3}\\ =\sqrt{669410.9}\\ =818.18\end{array}$

Therefore true speed of jet is $818.18\text{mi/h}$.

If ${\theta }_1$ is the angle made by true velocity vector p with the positive $x$ axis,

Angle ${\theta }_1$ has the property that,

$\tan {\theta }_1=\frac{\text{Vertical}\text{component}}{\text{Horizontal}\text{component}}$

From section (c), true velocity vector of  jet is $568.4i+588.5j$.

Thus,

$\begin{array}{c}\tan {\theta }_1=\frac{588.5}{568.4}\\ =1.04\end{array}$

Solve for ${\theta }_1$.

$\begin{array}{c}{\theta }_1={\tan }^{-1}\left(1.04\right)\\ =46.12°\end{array}$

So, direction is $\text{N}\left(90-46.12\right)°\text{E}$ or $\text{N}\text{43}\text{.88}°\text{E}$.

Therefore true speed of jet is $818.18\text{mi/h}$ and direction is $\text{N}\text{43}\text{.88}°\text{E}$.