Because the impulse is larger when the plane is taking off or landing
Explanation:
The impulse exerted on an object is equal to the change in momentum of the object:
[tex]I=\Delta p[/tex]
Where the change in momentum is given by
[tex]\Delta p = m \Delta v[/tex]
where
m is the mass of the object
[tex]\Delta v[/tex] is the change in velocity of the object
Let's now consider the airplane in the different situations:
- When the plane is flying at constant speed, the change in velocity is zero: [tex]\Delta v = 0[/tex], which means that the change in momentum is also zero: [tex]\Delta p = 0[/tex], and therefore, the impulse experienced is zero: [tex]I=0[/tex]. So, the ride is more comfortable
- When the plane is taking off or landing, the velocity of the plane is changing, so [tex]\Delta v \neq 0[/tex], which means that the change in momentum is also non-zero: [tex]\Delta p \neq 0[/tex], and therefore, the impulse experienced is also non-zero: [tex]I\neq 0[/tex]. This is the reason why the ride in this situation is less comfortable.
Learn more about impulse:
brainly.com/question/9484203
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