What Is the Resultant Velocity of an Airplane in Wind?

[v3ra]

An airplane has an air speed of 4.20 × 10^2 km/h [N 45° E]. The wind speed is 30 km/h
to the west.

a) What is the airplane’s resultant velocity?

I decided to use the trigonometric solution, but I am confused on how to break the equation down in order to get 399 km/h

c2 = a2 + b2 - 2ab.cosC
c2 = (4.20 x 10^2)^2 + (30)^2 - 2( 420) x (30)cos135
c2 = (17640000 + 900) - 17819
c2 = 17623081
c2 = ? (I tried square root of 17623081, but of course this is far off from the answer)

Please assist!

 

[szimmy]

You're on the right track with the trigonometric approach, but you went about it wrong. The initial velocity you have is 4.2 x 10^2 at an angle of 45 degrees North of East. Isn't your c term the 4.2 x 10^2?
From here you have a wind slowing down the x direction of the plane.
The velocity you have now is incorporating both the x and y directions. How can you divide the velocity up so you can deal with the x and y directions separately?

 

[v3ra]

Hmm.. I am not sure what you mean. I am terrible at this! I don`t know the velocity (the 339 km/h is the answer I am suppose to find) so how would I divide it up?

 

[mishek]

Are you familiar with vector addition?

 

[v3ra]

Yes, but I am not sure if I am doing it correctly.
Would a = 420 km/h [N 45 E] + b = 30 km/h [W] equals the resultant? But, when drawing it, at what angle would 30 km/h [W] be?

 

[AJ Bentley]

Draw it.

You will see it's a simple case of a 30 km/h headwind at 45 degrees off the plane's nose.

So the effect of the headwind along the path of the plane will be 30*cos (45). Subtract that from 420 and you have your answer.

You can use trig if you like - but a bit of thought will often save you having to remember complicated formulae.