Aplane is flying east at 135 m/s. the wind accelerates it at 2.18 m/s^2 directly northeast. after 18 s, what is the magnitude of the displacement of the plane?

When we say directly northeast that is equivalent to 45˚ north of east.

First let us determine the north and east components of the acceleration using cos and sin functions:

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

Then we set to determine the east component of the plane’s displacement by calculating using the formula:

d = vi * t + ½ * a * t^2 
d = 135 * 18 + ½ * 2.18 * cos 45 * 18^2 
d = 2430 + 353.16 * cos 45 = 2679.72 m

Calculating for the north component:
North = ½ * 2.18 * sin 45 * 18^2

North = 249.72 m 

Hence magnitude is:

Magnitude = sqrt (2679.72^2 + 249.72^2)

Magnitude = 2,691. 33 m

Calculating for angle:

Tan θ = North ÷ East 
Tan θ = 249.72 m  ÷ 2679.72 m

θ = 5.32°

So the plane was flying at 2,691. 33 m at 5.32°

it is called kinetic energy

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