A certain aircraft has a mass of 300,000 kg.At a certain instant during its landing,its speed is 27.0 m/s.If the stopping force exerted on the airplane is a constant 445,000 N,what is the speed of the airplane 10.0 s later?
A)10.0 m/s
B)12.2 m/s
C)14.0 m/s
D)18.0 m/s
E)20.0 m/s

Question

Quizplus · Accepted Answer

The deceleration can be calculated using Newton's second law: a = F/m = 445,000 N / 300,000 kg = 1.4833 m/s². The speed after 10 seconds is v = u + at = 27.0 m/s - 1.4833 m/s² * 10 s = 12.167 m/s, approximately 12.2 m/s.

A Certain Aircraft Has a Mass of 300,000 Kg