Chapter 9.1, Problem 59E

(a)

To determine

To express: The velocity of the wind as a vector in component form.

Answer to Problem 59E

The expression of wind velocity in component form is $40j$.

Explanation of Solution

Given:

The speed of wind is $40\text{mi/h}$ and wind is blowing in north direction.

Formula used:

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

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

Calculation:

The speed of wind is $40\text{mi/h}$ and direction of wind is north.

The angle made by wind velocity with positive $x$ axis is given by $\theta =90°$.

Wind velocity vector v is shown on coordinate axis in Figure (1).

Figure (1)

Substitute $40$ for $\left|v\right|$, $90°$ for $\theta$ in equation (1).

$\begin{array}{c}v=40\cos 90°+40\sin 90°\\ =\left(40\times 0\right)i+\left(40\times 1\right)j\\ =40j\end{array}$

Thus, the component form of wind velocity v is $40j$.

(b)

To determine

To express: The velocity of the jet relative to air in component form.

Answer to Problem 59E

Expression of jet velocity in component form relative to air is $425i$.

Explanation of Solution

Given:

The speed of Jet is $425\text{mi/h}$ and direction of Jet is east.

Formula used:

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

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

Calculation:

The speed of Jet is $\text{425}\text{mi/h}$ and direction of Jet is in east direction.

So, the angle made by Jet velocity vector with positive $x$ axis is $0°$.

Jet velocity vector is shown in Figure (2).

Figure (2)

Substitute $425$ for $\left|u\right|$, $0°$ for $\theta$ in equation (2).

$\begin{array}{c}u=425\cos 0°+425\sin 0°\\ =\left(425\times 1\right)i+\left(425\times 0\right)j\\ =425i\end{array}$

Thus, component form of Jet velocity u is $425i$.

(c)

To determine

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

Answer to Problem 59E

True velocity of jet as a vector is $425i+40j$.

Explanation of Solution

Given:

The speed of wind is $40\text{mi/h}$ and direction of wind is north.

The speed of Jet is $\text{425}\text{mi/h}$ and direction of Jet is East.

Formula used:

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

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

Calculation:

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

$p=u+v$

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

Figure (3)

From section (b) jet velocity u is $425i$ and from section(1) wind velocity v is $40j$.

Substitute $425$ for $a_1$, $0$ for $a_2$, $0$ for $b_1$ and $40$ for $b_2$ in equation (3).

$\begin{array}{c}u+v=\left(425+0\right)i+\left(0+40\right)j\\ =425i+40j\end{array}$

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

Therefore, $p=425i+40j$.

Thus true velocity of jet is $425i+40j$.

(d)

To determine

To find: True speed and direction of Jet.

Answer to Problem 59E

True speed of jet is $426.87\text{mi/h}$ and direction of jet is $\text{N}84.63°\text{E}$.

Explanation of Solution

Given:

The speed of Jet relative to air is $\text{425}\text{mi/h}$ and direction is east.

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 and $a_2$ is the vertical component.

Calculation:

From Section (c), the true velocity of jet is $425i+40j$.

Substitute $425$ for $a_1$ and $40$ for $a_2$ in equation (3).

$\begin{array}{l}\left|p\right|=\sqrt{{\left(425\right)}^2+{\left(40\right)}^2}\\ =\sqrt{180625+1600}\\ =\sqrt{182225}\\ =426.87\end{array}$

Therefore true speed of jet is $426.87\text{mi/h}$.

If ${\theta }_1$ is the angle made by true velocity vector p with the positive $x$ axis, then 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 $425i+40j$.

Thus,

$\begin{array}{c}\tan {\theta }_1=\frac{40}{425}\\ =0.094\end{array}$

Solve for ${\theta }_1$.

$\begin{array}{c}{\theta }_1={\tan }^{-1}\left(0.094\right)\\ =5.37°\end{array}$

So, direction is $\text{N}\left(90-5.37\right)°\text{E}$ or $\text{N}84.63°\text{E}$.

Therefore, the true speed of jet is $426.87\text{mi/h}$ and direction is $\text{N}84.63°\text{E}$.