Question

An elevator is initially moving upward at a speed of 12.00 m/s. the elevator experiences a constant downward acceleration of magnitude 4.00 m/s2 for 3.00 s. (a) find the magnitude and direction of the elevator’s final velocity. (b) how far did it move during the 3.00 s interval?

Answer 1

The two equations used in this problem is from the derived equations of rectilinear motion at a constant acceleration. 

a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance 
v is the final velocity
v₀ is the initial velocity
t is the time

Part a.) The acceleration is negative because it is moving downward: a = -4 m/s². The initial velocity is v₀ = 12 m/s and t is 3 s.

-4 = (v - 12)/3
v = 0 m/s
Thus, after 3 seconds the elevator comes to a stop.

Part b.) 

2(-4)x = 0² - 12²
x = 18 m
The elevator has traveled a distance of 18 m within 3 seconds.