Question

. An airplane's compass indicates that it is
headed due north, and its airspeed indicator
shows that it is moving through the air at 240
km/h. If there is a 100-km h wind from west to
east. In what direction should the pilot head to
travel due north? What will be her velocity
relative to the earth? 1
a. 35º W of N, 198 km/h
b. 28° W of N, 210 km/h
c. 25° W of N, 218 km/h
d. 29° W of N, 200 km/h​

Answer 1

Answer:

Therefore, the Wind Correction Angle will be (100*90)÷240 = 37.5 degrees.

Therefore, to maintain its track, the aircraft will have heading of 360–37.5 = 322.5 degrees.

Its speed relative to the Earth or Ground Speed will be 240 km/h.

Answer 2

Given:

An airplane's compass indicates that it is

headed due north, and its airspeed indicator

shows that it is moving through the air at 240

km/h. If there is a 100-km h wind from west to

east.

To find:

Direction in which pilot should travel so as to go North ; Velocity Of plane in stationary sky.

Calculation:

First, refer to the attached diagram to understand the vector.

Let actual velocity of plane be v in stationary sky.

So, as per triangle law of vector addition:

$\therefore \: {240}^{2} = {v}^{2} + {100}^{2}$

$= > \: 56700 = {v}^{2} + 10000$

$= > \: {v}^{2} = 46700$

$= > \: v = \sqrt{46700}$

$\boxed{ = > \: v = 216 \: km {hr}^{ - 1} }$

Let the angle with Y axis be $\theta$.

$\therefore \: \cos( \theta) = \dfrac{216}{240} = 0.9$

$= > \: \theta = { \cos}^{ - 1} (0.9)$

$= > \: \theta = 25 \degree$

So, the direction will be 25° West of North .