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A plane lands with a velocity of +126.1 m/s and decelerates at a maximum rate of 5.2 m/s^2?
A plane lands with a velocity of +126.1 m/s and decelerates at a maximum rate of 5.2 m/s^2.
From the instant the plane touches the runway, what is the minimum time needed before
it can come to rest?
Answer in units of s
The plane is landing on a naval aircraft carrier
that is 0.80 km long.
What distance does the plane require to
land?
Answer in units of km
I dont understand this.
2 Answers
- electron1Lv 78 years ago
vf = vi + a * t
vi = 126.1, vf = 0, a = -5.2 (The acceleration is negative, because the plane is decelerating)
0 = 126.1 – 5.2 * t
t = -126.1 ÷ -5.2 = 24.25 seconds
Use the following equation to determine then distance.
d = vi * t + ½ * t^2
d = 126.1 * 24.25 + ½ * -5.2 * 24.25^2
d = 1528.9625 meters
This is 1.5289625 km. This is longer than the aircraft carrier.
