Question

Velocity

The speed of an airplane$300mi/h$ is relative to the air. The wind is blowing due north with a speed of $30mi/h$. In what direction should the airplane head in order to arrive at a point due west of its location?

Short answer

The airplane should head in the direction $S84.26\text{W}$in order to arrive at a point due west of its location.

Step-by-step solution

Step 1. Given information

Given that the speed of an airplane is $300\text{mi}/\text{h}$ relative to the air. The wind is blowing due north with a speed of $30\text{mi}/\text{h}$.

Step 2. Concept used

Express all the velocities in the vector form. The component of the true velocity of the airplane in the north direction should be zero in order to arrive at a point due west of its location. Simplify for $\theta$ to find the direction.

Step 3. Calculation

Let,

$u=$velocity of wind

$v=$velocity of airplane relative to the air

The wind is blowing due north with a speed of $30\text{mi}/\text{h}$

Therefore, $u=30j$

The speed of an airplane is $300\text{mi}/\text{h}$relative to the air.

Therefore, $\left|v\right|=300$

Let the airplane make an angle $\theta$as shown in the figure.

We can write

$v=300\cos \theta i+300\sin \theta j$

The true velocity of the airplane will be

$\begin{array}{l}w=u+v\\ w=30j+300\cos \theta i+300\sin \theta j\\ w=300\cos \theta i+(30+300\sin \theta )j\end{array}$

In order to arrive at a point due west of its location, the airplane should have zero component of velocity in the vertical direction.

Therefore,

$\begin{array}{l}30+300\sin \theta =0\\ 300\sin \theta =-30\\ \sin \theta =-\frac{30}{300}\\ \sin \theta =-\frac{1}{10}\end{array}$