Question

A plane 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?

Answer 1

2,691.33 - magnitude

Step-by-step explanation:

5.32 - direction

Answer 2

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°