Respuesta :
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.
Answer:
[tex]v_w = 44.7 km/h[/tex]
[tex]\theta = 63.4 degree[/tex]
Explanation:
As we know that the velocity of the plane is
[tex]v_1 = 220 km/h[/tex] towards west
after half an hour the displacement of the plane in west direction is given as
[tex]x = 120 km[/tex]
towards south the displacement is given as
[tex]y = 20 km[/tex]
now we know that net velocity towards west
[tex]v_p + v_w = \frac{120}{0.5}[/tex]
[tex]220 + v_w = 240[/tex]
[tex]v_w = 20 km/h[/tex]
Similarly towards south we have
[tex]v_w = \frac{20}{0.5}[/tex]
[tex]v_w = 40 km/h[/tex]
So the net speed of the wind is given as
[tex]v_w = 20 \hat i + 40\hat j[/tex]
[tex]v_w= \sqrt{20^2 + 40^2}[/tex]
[tex]v_w = 44.7 km/h[/tex]
direction of wind is given as
[tex]tan\theta = \frac{v_y}{v_x}[/tex]
[tex]\theta = tan^{-1}\frac{40}{20}[/tex]
[tex]\theta = 63.4 degree[/tex]
