Chapter 9.1, Problem 62E

True Velocity of a Jet In what direction should the pilot in Exercise 60 head the plane for the true course to be due north?

To determine

To find: The direction of jet plane to head so that the true course is along north direction.

Answer to Problem 62E

The direction of jet plane should be $\text{N}\text{1}\text{.7}°\text{W}$.

Explanation of Solution

Given:

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

Calculation:

Vector u represents velocity of wind, vector v represents velocity of jet and vector p represents the resultant true velocity of jet plane which is in north direction, as shown below in 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,

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

And $\theta$ is,

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

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.

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

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

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

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

Substitute $765$ for $\left|v\right|$ in above formula.

$\begin{array}{c}{v}_1=\left|v\right|\cos \theta \\ =765\cos \theta \end{array}$

Vertical component $v_2$ of velocity vector of jet is,

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

Substitute $765$ for $\left|v\right|$ in above formula.

$\begin{array}{c}{v}_2=\left|v\right|\sin \theta \\ =765\sin \theta \end{array}$

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

Therefore, vector v is,

$\begin{array}{c}v=v_{1}i+v_{2}j\\ =765\cos \theta i+765\sin \theta j\end{array}$

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

$p=u+v$

Substitute $\left(27.5i+47.6j\right)$ for u and $\left(765\cos \theta i+765\sin \theta j\right)$ for v in above relation.

$\begin{array}{c}p=\left(27.5i+47.6j\right)+\left(765\cos \theta i+765\sin \theta j\right)\\ =\left(27.5+765\cos \theta \right)i+\left(47.6+765\sin \theta \right)\end{array}$

If the true course of the jet is in north direction, then horizontal component should be zero.

$27.5+765\cos \theta =0$

Solve for $\cos \theta$.

$\cos \theta =\frac{-27.5}{765}$

Solve for $\theta$.

$\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{-27.5}{765}\right)\\ =91.7°\end{array}$

Thus direction is $\text{N}\left(91.7-90\right)°\text{W}$ or $\text{N}\text{1}\text{.7}°\text{W}$

Thus, the direction of jet plane should be $\text{N}\text{1}\text{.7}°\text{W}$.