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

`v_w = 44.7 km/h`

`\theta = 63.4 degree`

Step-by-step 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 = (120)/(0.5)`

`220 + v_w = 240`

`v_w = 20 km/h`

Similarly towards south we have

`v_w = (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= √(20^2 + 40^2)`

`v_w = 44.7 km/h`

direction of wind is given as

`tan\theta = (v_y)/(v_x)`

`\theta = tan^(-1)(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.