We investigated a jet landing on an aircraft carrier. in a later maneuver, the jet comes in for a landing on solid ground with a speed of 110 m/s, and its acceleration can have a maximum magnitude of 7.42 m/s2 as it comes to rest. (a) from the instant the jet touches the runway, what is the minimum time interval needed before it can come to rest? 14.8 s (b) can this jet land on a small tropical island airport where the runway is 0.800 km long? yes

Time  is 14.8 s and cannot landing

Explanation:

This is a kinematic exercise with constant acceleration, we assume that the acceleration of the jet to take off and landing  are the same

Calculate the time to stop, where it has zero speed

       Vf² = Vo² + a t

       t = - Vo² / a

       t = - 110²/(-7.42)

       t = 14.8 s

This is the time it takes to stop the jet

Let's analyze the case of the landing at the small airport, let's look for the distance traveled to land, where the speed is zero

Vf² = Vo² + 2 to X

X = -Vo² / 2 a

X = -110² / 2 (-7.42)

X = 815.4 m

Since this distance is greater than the length of the runway, the jet cannot stop

Let's calculate the speed you should have to stop on a track of this size

Vo² = 2 a X

Vo = √ (2 7.42 800)

Vo = 109 m / s

It is conclusion the jet must lose some speed to land on this track

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