A jet lands on an aircraft carrier at 140 mi/h (≈63 m/s).

(A) What is its acceleration (assumed constant) if it stops in 2.0 s due to an arresting cable that snags the airplane and brings it to a stop?

(B) If the plane touches down at position $x_{i}=0$ , what is the final position of the plane?

Question

A jet lands on an aircraft carrier at 140 mi/h (≈63 m/s).

(A) What is its acceleration (assumed constant) if it stops in 2.0 s due to an arresting cable that snags the airplane and brings it to a stop?

(B) If the plane touches down at position [latex] x_{i}=0 [/latex], what is the final position of the plane?

Accepted Answer

We define our x axis as the direction of motion of the jet. A careful reading of the problem reveals that in addition to being given the initial speed of 63 m/s, we also know that the final speed is zero. We also note that we have no information about the change in position of the jet while it is slowing down. Equation 2.9 is the only equation in Table 2.2 that does not involve position, and so we use it to find the acceleration of the jet, modeled as a particle:

[latex] v_{x f}=v_{x i}+a_{x} t \quad\left( ext { for constant } a_{x} ight) [/latex] (2.9).

Table 2.2 Kinematic Equations for Motion of a Particle Under Constant Acceleration
Information Given by Equation	Equation
Velocity as a function of time	[latex] v_{x f}=v_{x i}+a_{x} t [/latex]
Position as a function of velocity and time	[latex] x_{f}=x_{i}+\frac{1}{2}\left(v_{x i}+v_{x f} ight) t [/latex]
Position as a function of time	[latex] x_{f}=x_{i}+v_{x i} t+\frac{1}{2} a_{x} t^{2} [/latex]
Velocity as a function of position	[latex] v_{x f}^{2}=v_{x i}^{2}+2 a_{x}\left(x_{f}-x_{i} ight) [/latex]

[latex] a_{x}=\frac{v_{x f}-v_{x i}}{t} \approx \frac{0-63 m / s }{2.0 s } [/latex].

[latex] =-31 m / s ^{2} [/latex].

(B) If the plane touches down at position [latex] x_{i}=0 [/latex], what is the final position of the plane?

We can now use any of the other three equations in Table 2.2 to solve for the final position. Let us choose Equation 2.11:

[latex] x_{f}=x_{i}+\frac{1}{2}\left(v_{x i}+v_{x f} ight) t=0+\frac{1}{2}(63 m / s +0)(2.0 s ) [/latex].

= 63 m.

[latex] x_{f}=x_{i}+\frac{1}{2}\left(v_{x i}+v_{x f} ight) t \quad ext { (for constant } a_{x} ext { ) } [/latex] (2.11)

If the plane travels much farther than this, it might fall into the ocean. The idea of using arresting cables to slow down landing aircraft and enable them to land safely on ships originated at about the time of the first World War. The cables are still a vital part of the operation of modern aircraft carriers.

Question 2.7: A jet lands on an aircraft carrier at 140 mi/h (≈63 m/s). (A...

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