Suppose the plane lands on the deck of the aircraft carrier with a speed higher than 63 m/s but with the same acceleration as that calculated in part (A). How will that change the answer to part (B)?

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If the plane is traveling faster at the beginning, it will stop farther away from its starting point, so the answer to part (B) should be larger. Mathematically, we see in Equation
2.11 that if [latex] v_{x i} ext { is larger, then } x_{f} ext { will be larger. } [/latex]

[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 landing deck has a length of 75 m, we can find the maximum initial speed with which the plane can land and still come to rest on the deck at the given acceleration from Equation 2.13:

[latex] v_{x f}^{2}=v_{x i}{ }^{2}+2 a_{x}\left(x_{f}-x_{i} ight) \quad ext { (for constant } a_{x} ext { ) } [/latex] (2.13).

[latex] v_{x f}^{2}=v_{x i}{ }^{2}+2 a_{x}\left(x_{f}-x_{i} ight) [/latex].

[latex] ightarrow v_{x i}=\sqrt{v_{x f}^{2}-2 a_{x}\left(x_{f}-x_{i} ight)} [/latex].

[latex] =\sqrt{0-2\left(-31 m / s ^{2} ight)(75 m -0)} [/latex].

= 68 m/s.

Question 2.W I .7: Suppose the plane lands on the deck of the aircraft carrier ...

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