Question

a plane 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?

Answer 1

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.

Answer 2

The answer for your question is 2680