Q38E - University Physics with Modern Physics

Question

An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50.0mi/h) is blowing toward the south. (a) If the airspeed of the plane (its speed in still air) is 320.0 km/h (about 200.0 mi/h), in which direction should the pilot head? (b) What is the speed of the plane over the ground? Draw a vector diagram.

Step-by-Step Solution

Answer

1. The plane will head towards northwest direction at 14^o.
2. The velocity of plane relative to ground is 310 km/h.

Step 1: Identification of given data

The given data can be listed below,

• The velocity of air is, $v_{AG}=80km/h$
• The airspeed of plane is, $v_{PA}=320km/h$

Step 2: Concept/Significance of relative velocity.

The speed or velocity of the object with respect to a stationary or traveling object or person is known as its relative velocity.

Step 3: (a) Determination of the direction of planein which the pilot should head

The vector diagram for velocity vectors is shown as below,

From diagram, the direction of plane is calculated as,

$\sin \theta =\frac{v_{AG}}{v_{PA}}$

Here, v_AG is the speed of air and v_PA is the speed of the plane.

Substitute all the values in the above,

$\begin{array}{rcl}\sin \theta & =& \frac{80km/h}{320km/h}\\ \theta & =& {\sin }^{-1}\left(\frac{1}{4}\right)\\ & =& {14}^o\end{array}$

Thus, the plane will head towards northwest direction at 14^o.

Step 4: (b) Determination of the speed of the plane over the ground.

BY, Pythagoras theorem, the velocity of the plane relative to ground is given by,

$v_{PG}=\sqrt{{\left(v_{PA}\right)}^2-{\left(v_{AG}\right)}^2}$

Here, v_AG is the speed of air and v_PA is the speed of the plane.

Substitute all the values in the above,

$\begin{array}{rcl}{v}_{PG}& =& \sqrt{{\left(320km/h\right)}^2-{\left(80km/h\right)}^2}\\ & =& \sqrt{\left(102400\right)-\left(6400\right)}km/h\\ & =& \sqrt{96000}km/h\\ & =& 310km/h\end{array}$

Thus, the velocity of plane relative to ground is 310km/h.