Question

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 $N30°W$

Short answer

The true speed of the jet will be$793.93\text{mi}/\text{h}$ in the direction $N{25.56}^{°}\text{W}$.

Step-by-step solution

Step 1. Given information

Given that a jet is flying through a wind that is blowing with a speed of $55\text{mi}/\text{h}$ in the direction $N{30}^{°}E$. The jet has a speed of $765\text{mi}/\text{h}$ relative to the air.

Step 2. Concept used

Write the velocity of the jet relative to the air and the velocity of the air in the vector form. Use formula to find the magnitude of the velocity,

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

Find the smallest positive angle $\theta$ between the positive x -axis and the vector to find the direction of the jet.

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

Simplify to get the answer in the vector form.

Step 3. Calculation

$u=$velocity of the wind

$v=$velocity of the jet relative to the air

The wind is blowing with a speed of $55\text{mi}/\text{h}$in the direction $N{30}^{°}\text{E}$

$\left|u\right|=55$

$\theta ={60}^{°}$from the east direction.

$\begin{array}{l}u=\left|u\right|\cos \theta i+\left|u\right|\sin \theta j\\ u=55\cos 60i+55\sin 60j\\ u=55\times \frac{1}{2}i+55\times \frac{\sqrt{3}}{2}j\\ u=27.5i+27.5\sqrt{3}j\end{array}$

The jet has a speed of $765\text{mi}/\text{h}$relative to the air in the direction $N{30}^{°}W$

$\left|v\right|=765$

$\theta ={120}^{°}$from the positive x axis.

$\begin{array}{l}v=\left|v\right|\cos \theta i+\left|v\right|\sin \theta j\\ v=765\cos 120i+765\sin 120j\\ v=765\left(\frac{-1}{2}\right)i+765\left(\frac{\sqrt{3}}{2}\right)j\\ v=-382.5i+382.5\sqrt{3}j\end{array}$

$\begin{array}{l}w=v+u\\ w=27.5i+27.5\sqrt{3}j-382.5i+382.5\sqrt{3}j\\ w=27.5i-382.5i+27.5\sqrt{3}j+382.5\sqrt{3}j\\ w=-355i+410\sqrt{3}j\end{array}$

From the above figure, we see that the direction $\theta$ has the property that,

$\begin{array}{l}\tan \theta =-\frac{410\sqrt{3}}{355}\\ \tan \theta =-\frac{410\sqrt{3}}{355}\\ \tan \theta =-2\end{array}$

Since the terminal point of the vector lies in the second quadrant, it follows $\theta ={116.56}^{°}$

Or $N{25.56}^{°}\text{W}$