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An elevator is initially moving upward at a speed of 12.72 m/s...Kinematics Question.?
An elevator is initially moving upward at a speed of 12.72 m/s. The elevator experiences a constant downward acceleration of magnitude 4.40 m/s2 for 2.91 s.
(a)Find the magnitude and direction of the elevator's final velocity.
(b)How far did it move during the 2.91 s interval?
1 Answer
- SteveLv 710 years agoFavorite Answer
a) Vf = Vi +a*t = 12.72 - 4.40*2.91 = -.084 m/s
b)
y1 = Vi²/2a = 18.386 m; t2 = t - t1 = 2.91 - Vf/a = .0191 sec
y2 = t2*Vf/2 = .0008 m
D = y1 + y2 = 18.3868 m
