Aplane is flying east at 135 m/s. the wind accelerates it at 2.18 m/s^2 directly northeast. after 18.0s, what is the magnitude of the displacement of the plane?

Directly northeast is 45˚ north of east. Let’s determine the north and east components of the acceleration. 

North = 2.18 * sin 45 
East = 2.18 * cos 45 

Both of these are approximately 1.54 m/s^2. Let’s use the following to determine the east component of the plane’s displacement. 

d = vi * t + ½ * a * t^2 
d = 135 * 18 + ½ * 2.18 * cos 45 * 18^2 
d = 2430 + 353.16 * cos 45 
This is approximately 2680 meters. 

North = ½ * 2.18 * sin 45 * 18^2 = 353.16 * sin 45 

This is approximately 250 meters. To determine the angle north of east, use the following equation. 

Tan θ = North ÷ East 
Tan θ = 353.16 * sin 45 ÷ (2430 + 353.16 * cos 45) 

This is approximately 5.3˚ north of east.

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