Question

An airplane touches down on the runway with a speed of 70 m/s2. Determine the airplane after each second of its deceleration.

Answer 1

vf^2 = vi^2 + 2*a*d
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vf = velocity final
vi = velocity initial
a = acceleration
d = distance
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since the airplane is decelerating to zero, vf = 0
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0 = 55*55 + 2*(-2.5)*d
d = (-55*55)/(2*(-2.5))
d = 605 meters