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February 1, 2015

February 1, 2015

Posted by **Physics** on Friday, September 18, 2009 at 9:18pm.

Question:

Find the air velocity of a plane that must have a realtive ground velocity of 250.0 h^-1 km norht if it encounters a wind pushing it toward the northeast at 75 h^-1 km

The Work:

I do not follow it one bit...

Va = Vg - Vw

= 250 h^-1 km north - 75.0 h^-1 km northeast

= 250.0 h^-1 km north + 75.0 h^-1 km southwest

Va^2 = Vg^2 + Vw^2 - 2VgVw cos A

= (250.0 h^-1 km)^2 + (75.0 h^-1 km)^2 - 2(250.0 h^-1 km)(75.0 h^-1 km)(cos 45.0)

= 62 500 h^-2 km^2 + 5 625 h^-2 km^2 + 26 516 h^-2 km62

= 41 609 h^-2 km^2

Va = 204 h^-1 km

Vw/sin W = Va sin A

sin W = Vw sin A/Va

= (75 h^-1 km)(sin 45.0)/(204 h^-1 km)

= (75 h^-1 km)(.707)/(204 h^-1 km)

= 0.260

B = 15.1

Answer = 204 h^-1 km, 15.1 degrees west of north

ok i was lost on line 2 and three were the added and subtracted 75 with two different directions why are there two

line 3 were did this come from

- 2VgVw cos A

thanks!

- Physics -
**MathMate**, Friday, September 18, 2009 at 9:35pmPythagoras theorem applies to the sides adjacent to the right angle:

a²+b²=c²

The extension when the angle is not a right angle is called the cosine rule, which states that:

a²+b²-2ab cos(C) = c²

which is what line three is doing, the angle between NE and N being 45°.

The resultant (air velocity) was calculated at 204 km/hr.

The line

Vw/sin W = Va sin A

should read

Vw/sin W = Va/sin A

which is the application of the sine rule to find the angle of a triangle in which one side and the opposite angle is known, as well as the side opposite to the angle sought.

I have not checked the numerical calculations. You can do the checking as an exercise.

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