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kari74 [83]
3 years ago
6

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

Physics
1 answer:
ivann1987 [24]3 years ago
5 0
<span>vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
---
0 = 55*55 + 2*(-2.5)*d
d = (-55*55)/(2*(-2.5))
d = 605 meters


</span>
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