### Question Description

An airplane must fly at a ground speed of 425 km/h in a direction of 10.0 degrees east of south to be on course and on schedule. If the wind velocity is 25.0 km/h 40.0 degrees east of north. you have to use the resultant to correct the pilot's course

## Final Answer

425 km/hr=425,000m/hr

425,000m/hr=118.1 m/s

V(x)=sin 10*118.1=20.5 m/s in the x direction

V(y)=cos 10*118.1=116.3 m/s in the y direction

this is what the x and y components should be if the plane was flying at the planned speed and heading.

Now we have to find the components of the wind

25 km/hr=25,000 m/hr

25,000 m/hr=6.9 m/s

6.9m/s at a heading of 40 degrees east of north that is the wind speed

now we have to break it down into its components

V(x)=sin 40*6.9=4.4 m/s in the x direction

V(y)=cos 40*6.9=5.3m/s in the y direction

now when we add these vectors it should equal the planes planned speed and heading

V( x total)=20.5 m/s

20.5 m/s=4.4+x

20.5-4.4=x

16.1 m/s=x

The plane must fly that fast in the x direction

V(y total)=116.3

116.3=5.3+y

116.3-5.3=y

111 m/s=y

111 m/s=y

16.1 m/s=x

we can use pythagorean theorem to find the hypotenuse

111^2+16.1^2=h^2

112.2=h

y=111 m/s

x=16.1 m/s

h=112.2 m/s

the plane must fly about 8.5 degrees east of north

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