The acceleration of a descending airplane

[deveny7]

Homework Statement

An airplane descends in a circular pattern with a constant radius of 250 meters. The airplane has a horizontal speed of 75 m/s (constant) and a downward
speed of 5 m/s, which is increasing at a rate of 2 m/s²

Determine the acceleration of the airplane.

Homework Equations

a_n = v² / ρ

a = √(a_t² + a_n²)

The Attempt at a Solution

Finding the acceleration for a circular motion is easy, but I am having trouble including the downward acceleration. Any help is greatly appreciated!

 

[Ibix]

If all else fails, derive it from first principles.

Assuming that the aircraft is at height h and is circling the origin, can you write down its position as a function of time? If so, what's its acceleration?

 

[deveny7]

What equation could I use to include the height and downward speed as a function of time?

 

[rcgldr]

Finding the acceleration for a circular motion is easy, but I am having trouble including the downward acceleration.

You figured out the centripetal accleration, and the problem states the downwards acceleration is 2 m / s², which would be the components of the acceleration vector (horizontal and vertical). To get the magnitude, take the square root of the sum of the squares of the components.

 

[Nickie]

The acceleration of the airplane is a combination of its horizontal acceleration and its vertical acceleration. The horizontal acceleration remains constant at 75 m/s, while the vertical acceleration is increasing at a rate of 2 m/s2. Therefore, the total acceleration can be calculated using the Pythagorean theorem: a = √(at2 + an2), where at is the horizontal acceleration and an is the vertical acceleration.

Substituting the given values, we get:

a = √((75 m/s)2 + (5 m/s + 2 m/s2t)2)

= √(5625 m2/s2 + (5 m/s + 2 m/s2t)2)

= √(5625 m2/s2 + 25 m2/s2 + 20 m/s2t + 4 m2/s4t2)

= √(5650 m2/s2 + 20 m/s2t + 4 m2/s4t2)

= √(5650 m2/s2 + 20 m/s2(0) + 4 m2/s4(0)2) [since the airplane is already at a constant horizontal speed and is not changing in time]

= √(5650 m2/s2)

= 75.166 m/s2

Therefore, the acceleration of the airplane is approximately 75.166 m/s2. This can also be represented as 75.166 m/s2 in the direction of the resultant velocity vector, which is at an angle of 90 degrees with respect to the horizontal velocity vector.