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→ A) What is the off speed for a plane with an acceleration of 4.2 m/s2 that can take off after traveling 300 yds? ► B) A tri
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Answer #1

A)

Given:

a=4.2\ \text{m/s}^2\\ S=300\ \text{yds}=274\ \text m

Since it is a constant accelerated motion, the final velocity during take off is,

v=\sqrt{2aS}\\ =\sqrt{2\times4.2\times300}\ \text{m/s}\\ =48\ \text{m/s}

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