Question

An airplane pilot sets a compass course due west and maintains a speed of 220 km/h. after flying for a half hour she finds herself over a town 120 km west and 20 km south of her starting point. (a) find the wind velocity (magnitude and direction). (b) if the wind velocity is 40 km/h due south, in what direction should the pilot set her course to travel due west? use the same airspeed of 220 km/h.

Answer 1

Answer:

$v_w = 44.7 km/h$

$\theta = 63.4 degree$

Explanation:

As we know that the velocity of the plane is

$v_1 = 220 km/h$ towards west

after half an hour the displacement of the plane in west direction is given as

$x = 120 km$

towards south the displacement is given as

$y = 20 km$

now we know that net velocity towards west

$v_p + v_w = \frac{120}{0.5}$

$220 + v_w = 240$

$v_w = 20 km/h$

Similarly towards south we have

$v_w = \frac{20}{0.5}$

$v_w = 40 km/h$

So the net speed of the wind is given as

$v_w = 20 \hat i + 40\hat j$

$v_w= \sqrt{20^2 + 40^2}$

$v_w = 44.7 km/h$

direction of wind is given as

$tan\theta = \frac{v_y}{v_x}$

$\theta = tan^{-1}\frac{40}{20}$

$\theta = 63.4 degree$

Answer 2

The net speed due west is = distance traveled in west / time taken = 120/0.5 = 240 km/h.  

so airspeed due west is  = net speed - speed of plane = 240-220= 20 km/h.  

airspeed due south is = distance traveled in west / time taken= 20/0.5= 40 km/h.  

the magnitude of the wind velocity = √[(airspeed due south )² + (airspeed due west)²] = √ ( 40^2 + 20^2 ) = 44.72 km/h

the angle of airspeed south of west is tan⁻¹ ( airspeed due south / airspeed due west )= tan⁻¹(40/20)=63.43 degrees.  

if wind velocity is 40 km/h due south, her velocity should have 20 km/h component in north.

so component west = sqrt ( 220^2 - 40^2 ) = 216.33 km/h.  

the angle north of west is arctan( 40/216.33 ) = 10.47 degrees.