Chapter 9.1, Problem 61E

True Velocity of a Jet Find the true speed and direction of the jet in Exercise 60 if the pilot heads the plane in the direction N 30° W.

To determine

The true speed and direction of the jet plane.

Answer to Problem 61E

The true speed of the jet plane is $794\text{mi/h}$ and direction of the jet plane is $\text{N}\text{26}\text{.6}°\text{W}$.

Explanation of Solution

Given:

The velocity of wind is $55\text{mi/h}$ in the direction $\text{N}\text{30}°\text{E}$ and the velocity of jet is $765\text{mi/h}$ in the direction $\text{N}\text{30}°\text{W}$.

Formula used:

Magnitude of a vector $p=a_{1}i+a_{2}j$ is,

$\left|p\right|=\sqrt{a_1{}^2+a_2{}^2}$ (1)

Where, $a_1$ is horizontal component and $a_2$ is vertical component.

Calculation:

Vector u represents velocity of wind, vector v represents velocity of jet and vector p represents the resultant true velocity of jet plane as shown in figure (1).

Figure (1)

If $\theta$ is the angle made by vector u with the positive horizontal direction then, horizontal component $u_1$ of velocity of wind u is,

$u_1=\left|u\right|\cos \theta$

From Figure (1), value of $\theta$ is,

$\begin{array}{c}\theta =90°-30°\\ \theta =60°\end{array}$

Substitute 55 for $\left|u\right|$, $60°$ for $\theta$ in formula to calculate $u_1$.

$\begin{array}{c}{u}_1=\left|u\right|\cos \theta \\ =55\times \cos 60°\\ =55\times \frac{1}{2}\\ =27.5\end{array}$

Vertical component of velocity of wind is,

$u_2=\left|u\right|\sin \theta$

Substitute 55 for $\left|u\right|$, $60°$ for $\theta$ in formula to calculate $u_2$.

$\begin{array}{c}{u}_2=\left|u\right|\sin \theta \\ =55\times \sin 60°\\ =55\times \frac{\sqrt{3}}{2}\\ =47.6\end{array}$

Velocity vector u of wind is given by $u=u_{1}i+u_{2}j$.

Substitute $27.5$ for $u_1$ and $47.6$ for $u_2$ in above relation.

$\begin{array}{c}u=u_{1}i+u_{2}j\\ =27.5i+47.6j\end{array}$

The vector u is $u=27.5i+47.6j$.

If $\theta$ is the angle made by vector v with the positive horizontal direction then,

$\begin{array}{c}\theta =90°+30°\\ =120°\end{array}$

Horizontal component $v_1$ of velocity vector of jet is,

$v_1=\left|v\right|\cos \theta$

Substitute $765$ for $\left|v\right|$, $120°$ for $\theta$ in above formula.

$\begin{array}{c}{v}_1=\left|v\right|\cos \theta \\ =765\times \cos \left(120\right)°\\ =-382.5\end{array}$

Vertical component $v_2$ of velocity vector of jet is

$v_2=\left|v\right|\sin \theta$

Substitute $765$ for $\left|v\right|$, $120°$ for $\theta$ in above formula.

$\begin{array}{c}{v}_2=\left|v\right|\sin \theta \\ =765\times \sin \left(120°\right)\\ =662.5\end{array}$

Velocity vector v of river is given by $v=v_{1}i+v_{2}j$.

Substitute for $v_1$ and $662.5$ for $v_2$ in above relation.

$\begin{array}{c}v=v_{1}i+v_{2}j\\ =-382.5i+662.5j\end{array}$

If p represents the resultant true velocity vector of jet then,

$p=u+v$

Substitute $\left(27.5i+47.6j\right)$ for u and $\left(-382.5i+662.5j\right)$ for v in above relation.

$\begin{array}{l}p=\left(27.5i+47.6j\right)+\left(-382.5i+662.5j\right)\\ =\left(27.5-382.5\right)i+\left(47.6+662.5\right)j\\ =-355i+710.1j\end{array}$

Thus, true velocity of swimmer as a vector is $-355i+710.1j$.

Magnitude of true velocity of jet plane is,

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

Substitute $-355$ for $a_1$ and $710.1$ for $a_2$ in above formula.

$\begin{array}{c}\left|p\right|=\sqrt{{\left(-355\right)}^2+{\left(710.1\right)}^2}\\ =\sqrt{126025+504242.01}\\ =\sqrt{630267.01}\\ =794\end{array}$

Thus, true speed of jet is $794\text{mi/h}$.

The direction of true velocity of jet is given by ${\theta }_1$.

The angle $\theta$ has the property that,

$\tan {\theta }_1=\frac{\text{coefficient}\text{of}j}{\text{coefficient}\text{of}i}$

Vector of true velocity is $-355i+710.1j$.

Substitute $710.1$ for coefficient of j and $-355$ for coefficient of i in above relation.

$\begin{array}{c}\tan {\theta }_1=\frac{710.1}{-355}\\ =-2\end{array}$

Solve for $\theta$.

$\begin{array}{c}{\theta }_1={\tan }^{-1}\left(-2\right)\\ =-63.43°\end{array}$

Therefore, direction is,

$\begin{array}{c}{\theta }_1=90°-63.43°\\ =26.6°\end{array}$

The direction of jet is $\text{N}\text{26}\text{.6}°\text{W}$.

Thus, true speed of jet is $794\text{mi/h}$ and the direction of jet is $\text{N}\text{26}\text{.6}°\text{W}$.